Evaluate $sumlimits_{n=0}^{infty}(-1)^nsumlimits_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}$
$begingroup$
Evaluate $g(k)$ if $sumlimits_{n=0}^{infty}(-1)^nsumlimits_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)$.
The well-known Gregory Series,
$$sum_{n=0}^{infty}frac{(-1)^n}{2n+1}=frac{pi}{4}tag1$$
Let us generalize $(1)$ in terms of binomial coefficients ${k choose j}$,
Where $kge0$
$$sum_{n=0}^{infty}(-1)^nsum_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)tag2$$
Expand $(2)$ for $k=1,2$ and $3$,
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{1}{2n+3}right)=frac{pi}{2}-1tag3$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-2frac{1}{2n+3}+frac{1}{2n+5}right)=pi-frac{8}{3}tag4$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{3}{2n+3}+frac{3}{2n+5}-frac{1}{2n+5}right)=2pi-frac{88}{15}tag5$$
The pattern for $g(k)$ is not so obvious, I was not able to determine the closed form $g(k)$
Q:
How can we find the closed form for $(2)?$
sequences-and-series
edited Jan 1 at 17:35
Did
248k23224463
asked Dec 31 '18 at 16:32
user583851user583851
49919
$endgroup$
add a comment |
$begingroup$
Evaluate $g(k)$ if $sumlimits_{n=0}^{infty}(-1)^nsumlimits_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)$.
The well-known Gregory Series,
$$sum_{n=0}^{infty}frac{(-1)^n}{2n+1}=frac{pi}{4}tag1$$
Let us generalize $(1)$ in terms of binomial coefficients ${k choose j}$,
Where $kge0$
$$sum_{n=0}^{infty}(-1)^nsum_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)tag2$$
Expand $(2)$ for $k=1,2$ and $3$,
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{1}{2n+3}right)=frac{pi}{2}-1tag3$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-2frac{1}{2n+3}+frac{1}{2n+5}right)=pi-frac{8}{3}tag4$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{3}{2n+3}+frac{3}{2n+5}-frac{1}{2n+5}right)=2pi-frac{88}{15}tag5$$
The pattern for $g(k)$ is not so obvious, I was not able to determine the closed form $g(k)$
Q:
How can we find the closed form for $(2)?$
sequences-and-series
edited Jan 1 at 17:35
Did
248k23224463
asked Dec 31 '18 at 16:32
user583851user583851
49919
$endgroup$
add a comment |
$begingroup$
Evaluate $g(k)$ if $sumlimits_{n=0}^{infty}(-1)^nsumlimits_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)$.
The well-known Gregory Series,
$$sum_{n=0}^{infty}frac{(-1)^n}{2n+1}=frac{pi}{4}tag1$$
Let us generalize $(1)$ in terms of binomial coefficients ${k choose j}$,
Where $kge0$
$$sum_{n=0}^{infty}(-1)^nsum_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)tag2$$
Expand $(2)$ for $k=1,2$ and $3$,
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{1}{2n+3}right)=frac{pi}{2}-1tag3$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-2frac{1}{2n+3}+frac{1}{2n+5}right)=pi-frac{8}{3}tag4$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{3}{2n+3}+frac{3}{2n+5}-frac{1}{2n+5}right)=2pi-frac{88}{15}tag5$$
The pattern for $g(k)$ is not so obvious, I was not able to determine the closed form $g(k)$
Q:
How can we find the closed form for $(2)?$
sequences-and-series
edited Jan 1 at 17:35
Did
248k23224463
asked Dec 31 '18 at 16:32
user583851user583851
49919
$endgroup$
Evaluate $g(k)$ if $sumlimits_{n=0}^{infty}(-1)^nsumlimits_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)$.
The well-known Gregory Series,
$$sum_{n=0}^{infty}frac{(-1)^n}{2n+1}=frac{pi}{4}tag1$$
Let us generalize $(1)$ in terms of binomial coefficients ${k choose j}$,
Where $kge0$
$$sum_{n=0}^{infty}(-1)^nsum_{j=0}^{k}{k choose j}frac{(-1)^j}{2n+2j+1}=frac{pi}{2^{2-k}}+g(k)tag2$$
Expand $(2)$ for $k=1,2$ and $3$,
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{1}{2n+3}right)=frac{pi}{2}-1tag3$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-2frac{1}{2n+3}+frac{1}{2n+5}right)=pi-frac{8}{3}tag4$$
$$sum_{n=0}^{infty}(-1)^nleft(frac{1}{2n+1}-frac{3}{2n+3}+frac{3}{2n+5}-frac{1}{2n+5}right)=2pi-frac{88}{15}tag5$$
The pattern for $g(k)$ is not so obvious, I was not able to determine the closed form $g(k)$
Q:
How can we find the closed form for $(2)?$
sequences-and-series
sequences-and-series
edited Jan 1 at 17:35
Did
248k23224463
asked Dec 31 '18 at 16:32
user583851user583851
49919
edited Jan 1 at 17:35
Did
248k23224463
asked Dec 31 '18 at 16:32
user583851user583851
49919
edited Jan 1 at 17:35
Did
248k23224463
edited Jan 1 at 17:35
Did
248k23224463
edited Jan 1 at 17:35
Did
248k23224463
248k23224463
asked Dec 31 '18 at 16:32
user583851user583851
49919
asked Dec 31 '18 at 16:32
user583851user583851
49919
asked Dec 31 '18 at 16:32
user583851user583851
49919
49919
add a comment |
add a comment |
3 Answers
3
active
oldest
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$begingroup$
We will make use of the identity
$$
sum_{k=0}^n(-1)^kbinom{n}{k}frac1{k+alpha}=frac{n!,Gamma(alpha)}{Gamma(n+1+alpha)}tag1
$$
which can be proven by induction on $n$.
The sum in the question is
$$
begin{align}
&sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{2n+2j+1}tag2\
&=frac12sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{n+j+frac12}tag3\
&=frac12sum_{n=0}^infty(-1)^nfrac{k!,Gamma!left(n+frac12right)}{Gamma!left(n+k+frac32right)}tag4\
&=frac12sum_{n=0}^infty(-1)^nint_0^1(1-t)^kt^{n-frac12},mathrm{d}ttag5\
&=frac12int_0^1frac{(1-t)^k}{(1+t)sqrt{t}},mathrm{d}ttag6\
&=int_0^1frac{left(1-t^2right)^k}{1+t^2},mathrm{d}ttag7\
&=int_0^1frac{left(2-left(1+t^2right)right)^k}{1+t^2},mathrm{d}ttag8\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}int_0^1left(1+t^2right)^{j-1},mathrm{d}tright)tag9\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}sum_{i=0}^{j-1}binom{j-1}{i}frac1{2i+1}right)tag{10}\
&=2^kleft(fracpi4-frac12sum_{j=0}^{k-1}left(-frac12right)^jsum_{m=1}^kbinom{k-m}{j}sum_{i=0}^{j}binom{j}{i}frac1{2i+1}right)tag{11}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=i}^{k-m}left(-frac12right)^jbinom{k-m}{i}binom{k-m-i}{j-i}frac1{2i+1}right)tag{12}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=0}^{k-m-i}left(-frac12right)^{j+i}binom{k-m}{i}binom{k-m-i}{j}frac1{2i+1}right)tag{13}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}left(-frac12right)^ibinom{k-m}{i}left(frac12right)^{k-m-i}frac1{2i+1}right)tag{14}\
&=2^kleft(fracpi4-frac14sum_{m=1}^ksum_{i=0}^{k-m}(-1)^ibinom{k-m}{i}left(frac12right)^{k-m}frac1{i+frac12}right)tag{15}\
&=2^kleft(fracpi4-frac14sum_{m=1}^kleft(frac12right)^{k-m}frac{(k-m)!,Gamma!left(frac12right)}{Gamma!left(k-m+frac32right)}right)tag{16}\
&=2^kleft(fracpi4-frac14sum_{m=0}^{k-1}left(frac12right)^mfrac{m!,Gamma!left(frac12right)}{Gamma!left(m+frac32right)}right)tag{17}\
&=2^kleft(fracpi4-frac12sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}right)tag{18}\
end{align}
$$
Explanation:
$phantom{1}(3)$: bring $frac12$ out front
$phantom{1}(4)$: apply $(1)$
$phantom{1}(5)$: apply the Beta Function
$phantom{1}(6)$: sum in $n$
$phantom{1}(7)$: substitute $tmapsto t^2$
$phantom{1}(8)$: $1-t^2=2-left(1+t^2right)$
$phantom{1}(9)$: apply the Binomial Theorem
$(10)$: apply the Binomial Theorem and integrate
$(11)$: $binom{k}{j}=sumlimits_{m=1}^kbinom{k-m}{j-1}$ and substitute $jmapsto j+1$
$(12)$: reorder summation and $binom{k-m}{j}binom{j}{i}=binom{k-m}{i}binom{k-m-i}{j-i}$
$(13)$: substitute $jmapsto j+i$
$(14)$: apply the Binomial Theorem
$(15)$: shift factors of $2$
$(16)$: apply $(1)$
$(17)$: substitute $mmapsto k-m$
$(18)$: rewrite the Gamma functions as factorials
Thus,
$$
bbox[5px,border:2px solid #C0A000]{g(k)=-2^{k-1}sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}}tag{19}
$$
$(4)$ says that asymptotically, the full sum from the question is $frac12sqrt{fracpi{k}}$, which vanishes as $ktoinfty$. Thus, the sum in $(18)$ must converge quickly to $fracpi2$. In fact, the sum in $(18)$ is the partial sum of the sum in this question whose limit is $fracpi2$.
answered Jan 1 at 14:22
robjohn♦robjohn
268k27308634
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$begingroup$
Note that the partial fractions decomposition of the inverse
of the Rising Factorial gives:
$$
eqalign{
& left( {x - 1} right)^{,underline {, - left( {m + 1} right),} } = {1 over {x^{,overline {,m + 1,} } }}
= {{Gamma (x)} over {Gamma (x + m + 1)}} = ;quad left| {;0 le {rm integer }, m} right.quad = cr
& = sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {left( {m - k} right)!k!}}} right)}
= sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {m!}}binom{m}{k}
} right)} cr}
$$
So
$$
eqalign{
& sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {2n + 2j + 1}}} = cr
& = {1 over 2}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {left( {n + 1/2 + j} right)}}} = cr
& = {{k!} over {2left( {n + 1/2} right)^{,overline {,k + 1,} } }} cr}
$$
We have then that the sum in $n$ is
$$
eqalign{
& S(k) = sumlimits_{0, le ,n} {left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right){1 over {2n + 2j + 1}}} } = cr
& = {1 over 2}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{k!} over {left( {n + 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,k + 1,} } }
over {left( {1 + n - 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,n - 1/2,} } }
over {left( {2 + k} right)^{,overline {,n - 1/2,} } }}} = cr
& = {{1^{,overline {, - 1/2,} } } over {2left( {k + 1} right)left( {2 + k} right)^{,overline {, - 1/2,} } }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {2 + k} right)} over {2left( {k + 1} right)Gamma left( 1 right)Gamma left( {3/2 + k} right) }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {1 + k} right)} over {2,Gamma left( {3/2 + k} right)}}{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) = cr
& = 2^{,2,k} {{left( {k!} right)^{,2} } over {left( {2k + 1} right)!}};;{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) cr}
$$
where in the mid of the derivation we have used the identity
$$
{{z^{,overline {,w,} } } over {left( {z + a} right)^{,overline {,w,} } }} = {{z^{,overline {,a,} } } over {left( {z + w} right)^{,overline {,a,} } }}
$$
and at the end the duplication formula for Gamma
$$
Gamma left( {2,z} right) = {{2^{,2,z - 1} } over {sqrt pi }}Gamma left( z right)Gamma left( {z + 1/2} right) = 2^{,2,z - 1} {{Gamma left( z right)Gamma left( {z + 1/2} right)} over {Gamma left( {1/2} right)}}
$$
Alternative Approach
Actually there is a more straight alternative aproach through the Digamma Function
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {4m + 2j + 1}} - {1 over {4m + 2j + 3}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {m + j/2 + 1/4}} - {1 over {m + j/2 + 3/4}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k cr
2j cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right)left( {psi left( {j + 5/4} right) - psi left( {j + 3/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)psi left( {j + 3/4} right) - left( matrix{
k cr
2j cr} right)psi left( {j + 1/4} right) - left( matrix{
k cr
2j + 1 cr} right)psi left( {j + 1 + 1/4} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right){1 over {j + 1/4}}} cr}
$$
Since we can write (re. for instance to this page)
$$
eqalign{
& psi left( {j + 3/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 3}}} + {pi over 2} - ln 8 - gamma
= psi left( {j + 3/4} right) - left( {psi left( {3/4} right) - left( {{pi over 2} - ln 8 - gamma } right)} right) cr
& psi left( {j + 1/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 1}}} - {pi over 2} - ln 8 - gamma
= psi left( {j + 1/4} right) - left( {psi left( {1/4} right) - left( { - {pi over 2} - ln 8 - gamma } right)} right) cr}
$$
we arrive to express the sum as
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 3} right)left( {4,l + 1} right)}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 2} right)^{,2} - 1}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} cr}
$$
Interesting the comparison of this expression with the one given by
@robjohn
answered Jan 1 at 17:31
G CabG Cab
19.7k31339
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add a comment |
$begingroup$
(2) is
$$frac11-frac23+frac25-frac27+cdots=-1+2left(frac11-frac13+frac15-cdotsright),$$
(3) is
$$frac11-frac33+frac45-frac47+cdots=-3+frac13+4left(frac11-frac13+frac15-cdotsright),$$
(4) is
$$frac11-frac43+frac75-frac87+cdots=-7+frac43-frac15+8left(frac11-frac13+frac15-cdotsright),$$
etc.
In general, one gets
$$2^kfracpi4-sum_{j=0}^{k-1}(-1)^jfrac1{2j+1}
left(2^k-sum_{i=0}^j{kchoose i}right).$$
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
105k1160133
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
We will make use of the identity
$$
sum_{k=0}^n(-1)^kbinom{n}{k}frac1{k+alpha}=frac{n!,Gamma(alpha)}{Gamma(n+1+alpha)}tag1
$$
which can be proven by induction on $n$.
The sum in the question is
$$
begin{align}
&sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{2n+2j+1}tag2\
&=frac12sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{n+j+frac12}tag3\
&=frac12sum_{n=0}^infty(-1)^nfrac{k!,Gamma!left(n+frac12right)}{Gamma!left(n+k+frac32right)}tag4\
&=frac12sum_{n=0}^infty(-1)^nint_0^1(1-t)^kt^{n-frac12},mathrm{d}ttag5\
&=frac12int_0^1frac{(1-t)^k}{(1+t)sqrt{t}},mathrm{d}ttag6\
&=int_0^1frac{left(1-t^2right)^k}{1+t^2},mathrm{d}ttag7\
&=int_0^1frac{left(2-left(1+t^2right)right)^k}{1+t^2},mathrm{d}ttag8\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}int_0^1left(1+t^2right)^{j-1},mathrm{d}tright)tag9\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}sum_{i=0}^{j-1}binom{j-1}{i}frac1{2i+1}right)tag{10}\
&=2^kleft(fracpi4-frac12sum_{j=0}^{k-1}left(-frac12right)^jsum_{m=1}^kbinom{k-m}{j}sum_{i=0}^{j}binom{j}{i}frac1{2i+1}right)tag{11}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=i}^{k-m}left(-frac12right)^jbinom{k-m}{i}binom{k-m-i}{j-i}frac1{2i+1}right)tag{12}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=0}^{k-m-i}left(-frac12right)^{j+i}binom{k-m}{i}binom{k-m-i}{j}frac1{2i+1}right)tag{13}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}left(-frac12right)^ibinom{k-m}{i}left(frac12right)^{k-m-i}frac1{2i+1}right)tag{14}\
&=2^kleft(fracpi4-frac14sum_{m=1}^ksum_{i=0}^{k-m}(-1)^ibinom{k-m}{i}left(frac12right)^{k-m}frac1{i+frac12}right)tag{15}\
&=2^kleft(fracpi4-frac14sum_{m=1}^kleft(frac12right)^{k-m}frac{(k-m)!,Gamma!left(frac12right)}{Gamma!left(k-m+frac32right)}right)tag{16}\
&=2^kleft(fracpi4-frac14sum_{m=0}^{k-1}left(frac12right)^mfrac{m!,Gamma!left(frac12right)}{Gamma!left(m+frac32right)}right)tag{17}\
&=2^kleft(fracpi4-frac12sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}right)tag{18}\
end{align}
$$
Explanation:
$phantom{1}(3)$: bring $frac12$ out front
$phantom{1}(4)$: apply $(1)$
$phantom{1}(5)$: apply the Beta Function
$phantom{1}(6)$: sum in $n$
$phantom{1}(7)$: substitute $tmapsto t^2$
$phantom{1}(8)$: $1-t^2=2-left(1+t^2right)$
$phantom{1}(9)$: apply the Binomial Theorem
$(10)$: apply the Binomial Theorem and integrate
$(11)$: $binom{k}{j}=sumlimits_{m=1}^kbinom{k-m}{j-1}$ and substitute $jmapsto j+1$
$(12)$: reorder summation and $binom{k-m}{j}binom{j}{i}=binom{k-m}{i}binom{k-m-i}{j-i}$
$(13)$: substitute $jmapsto j+i$
$(14)$: apply the Binomial Theorem
$(15)$: shift factors of $2$
$(16)$: apply $(1)$
$(17)$: substitute $mmapsto k-m$
$(18)$: rewrite the Gamma functions as factorials
Thus,
$$
bbox[5px,border:2px solid #C0A000]{g(k)=-2^{k-1}sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}}tag{19}
$$
$(4)$ says that asymptotically, the full sum from the question is $frac12sqrt{fracpi{k}}$, which vanishes as $ktoinfty$. Thus, the sum in $(18)$ must converge quickly to $fracpi2$. In fact, the sum in $(18)$ is the partial sum of the sum in this question whose limit is $fracpi2$.
answered Jan 1 at 14:22
robjohn♦robjohn
268k27308634
$endgroup$
add a comment |
$begingroup$
We will make use of the identity
$$
sum_{k=0}^n(-1)^kbinom{n}{k}frac1{k+alpha}=frac{n!,Gamma(alpha)}{Gamma(n+1+alpha)}tag1
$$
which can be proven by induction on $n$.
The sum in the question is
$$
begin{align}
&sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{2n+2j+1}tag2\
&=frac12sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{n+j+frac12}tag3\
&=frac12sum_{n=0}^infty(-1)^nfrac{k!,Gamma!left(n+frac12right)}{Gamma!left(n+k+frac32right)}tag4\
&=frac12sum_{n=0}^infty(-1)^nint_0^1(1-t)^kt^{n-frac12},mathrm{d}ttag5\
&=frac12int_0^1frac{(1-t)^k}{(1+t)sqrt{t}},mathrm{d}ttag6\
&=int_0^1frac{left(1-t^2right)^k}{1+t^2},mathrm{d}ttag7\
&=int_0^1frac{left(2-left(1+t^2right)right)^k}{1+t^2},mathrm{d}ttag8\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}int_0^1left(1+t^2right)^{j-1},mathrm{d}tright)tag9\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}sum_{i=0}^{j-1}binom{j-1}{i}frac1{2i+1}right)tag{10}\
&=2^kleft(fracpi4-frac12sum_{j=0}^{k-1}left(-frac12right)^jsum_{m=1}^kbinom{k-m}{j}sum_{i=0}^{j}binom{j}{i}frac1{2i+1}right)tag{11}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=i}^{k-m}left(-frac12right)^jbinom{k-m}{i}binom{k-m-i}{j-i}frac1{2i+1}right)tag{12}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=0}^{k-m-i}left(-frac12right)^{j+i}binom{k-m}{i}binom{k-m-i}{j}frac1{2i+1}right)tag{13}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}left(-frac12right)^ibinom{k-m}{i}left(frac12right)^{k-m-i}frac1{2i+1}right)tag{14}\
&=2^kleft(fracpi4-frac14sum_{m=1}^ksum_{i=0}^{k-m}(-1)^ibinom{k-m}{i}left(frac12right)^{k-m}frac1{i+frac12}right)tag{15}\
&=2^kleft(fracpi4-frac14sum_{m=1}^kleft(frac12right)^{k-m}frac{(k-m)!,Gamma!left(frac12right)}{Gamma!left(k-m+frac32right)}right)tag{16}\
&=2^kleft(fracpi4-frac14sum_{m=0}^{k-1}left(frac12right)^mfrac{m!,Gamma!left(frac12right)}{Gamma!left(m+frac32right)}right)tag{17}\
&=2^kleft(fracpi4-frac12sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}right)tag{18}\
end{align}
$$
Explanation:
$phantom{1}(3)$: bring $frac12$ out front
$phantom{1}(4)$: apply $(1)$
$phantom{1}(5)$: apply the Beta Function
$phantom{1}(6)$: sum in $n$
$phantom{1}(7)$: substitute $tmapsto t^2$
$phantom{1}(8)$: $1-t^2=2-left(1+t^2right)$
$phantom{1}(9)$: apply the Binomial Theorem
$(10)$: apply the Binomial Theorem and integrate
$(11)$: $binom{k}{j}=sumlimits_{m=1}^kbinom{k-m}{j-1}$ and substitute $jmapsto j+1$
$(12)$: reorder summation and $binom{k-m}{j}binom{j}{i}=binom{k-m}{i}binom{k-m-i}{j-i}$
$(13)$: substitute $jmapsto j+i$
$(14)$: apply the Binomial Theorem
$(15)$: shift factors of $2$
$(16)$: apply $(1)$
$(17)$: substitute $mmapsto k-m$
$(18)$: rewrite the Gamma functions as factorials
Thus,
$$
bbox[5px,border:2px solid #C0A000]{g(k)=-2^{k-1}sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}}tag{19}
$$
$(4)$ says that asymptotically, the full sum from the question is $frac12sqrt{fracpi{k}}$, which vanishes as $ktoinfty$. Thus, the sum in $(18)$ must converge quickly to $fracpi2$. In fact, the sum in $(18)$ is the partial sum of the sum in this question whose limit is $fracpi2$.
answered Jan 1 at 14:22
robjohn♦robjohn
268k27308634
$endgroup$
add a comment |
$begingroup$
We will make use of the identity
$$
sum_{k=0}^n(-1)^kbinom{n}{k}frac1{k+alpha}=frac{n!,Gamma(alpha)}{Gamma(n+1+alpha)}tag1
$$
which can be proven by induction on $n$.
The sum in the question is
$$
begin{align}
&sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{2n+2j+1}tag2\
&=frac12sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{n+j+frac12}tag3\
&=frac12sum_{n=0}^infty(-1)^nfrac{k!,Gamma!left(n+frac12right)}{Gamma!left(n+k+frac32right)}tag4\
&=frac12sum_{n=0}^infty(-1)^nint_0^1(1-t)^kt^{n-frac12},mathrm{d}ttag5\
&=frac12int_0^1frac{(1-t)^k}{(1+t)sqrt{t}},mathrm{d}ttag6\
&=int_0^1frac{left(1-t^2right)^k}{1+t^2},mathrm{d}ttag7\
&=int_0^1frac{left(2-left(1+t^2right)right)^k}{1+t^2},mathrm{d}ttag8\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}int_0^1left(1+t^2right)^{j-1},mathrm{d}tright)tag9\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}sum_{i=0}^{j-1}binom{j-1}{i}frac1{2i+1}right)tag{10}\
&=2^kleft(fracpi4-frac12sum_{j=0}^{k-1}left(-frac12right)^jsum_{m=1}^kbinom{k-m}{j}sum_{i=0}^{j}binom{j}{i}frac1{2i+1}right)tag{11}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=i}^{k-m}left(-frac12right)^jbinom{k-m}{i}binom{k-m-i}{j-i}frac1{2i+1}right)tag{12}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=0}^{k-m-i}left(-frac12right)^{j+i}binom{k-m}{i}binom{k-m-i}{j}frac1{2i+1}right)tag{13}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}left(-frac12right)^ibinom{k-m}{i}left(frac12right)^{k-m-i}frac1{2i+1}right)tag{14}\
&=2^kleft(fracpi4-frac14sum_{m=1}^ksum_{i=0}^{k-m}(-1)^ibinom{k-m}{i}left(frac12right)^{k-m}frac1{i+frac12}right)tag{15}\
&=2^kleft(fracpi4-frac14sum_{m=1}^kleft(frac12right)^{k-m}frac{(k-m)!,Gamma!left(frac12right)}{Gamma!left(k-m+frac32right)}right)tag{16}\
&=2^kleft(fracpi4-frac14sum_{m=0}^{k-1}left(frac12right)^mfrac{m!,Gamma!left(frac12right)}{Gamma!left(m+frac32right)}right)tag{17}\
&=2^kleft(fracpi4-frac12sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}right)tag{18}\
end{align}
$$
Explanation:
$phantom{1}(3)$: bring $frac12$ out front
$phantom{1}(4)$: apply $(1)$
$phantom{1}(5)$: apply the Beta Function
$phantom{1}(6)$: sum in $n$
$phantom{1}(7)$: substitute $tmapsto t^2$
$phantom{1}(8)$: $1-t^2=2-left(1+t^2right)$
$phantom{1}(9)$: apply the Binomial Theorem
$(10)$: apply the Binomial Theorem and integrate
$(11)$: $binom{k}{j}=sumlimits_{m=1}^kbinom{k-m}{j-1}$ and substitute $jmapsto j+1$
$(12)$: reorder summation and $binom{k-m}{j}binom{j}{i}=binom{k-m}{i}binom{k-m-i}{j-i}$
$(13)$: substitute $jmapsto j+i$
$(14)$: apply the Binomial Theorem
$(15)$: shift factors of $2$
$(16)$: apply $(1)$
$(17)$: substitute $mmapsto k-m$
$(18)$: rewrite the Gamma functions as factorials
Thus,
$$
bbox[5px,border:2px solid #C0A000]{g(k)=-2^{k-1}sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}}tag{19}
$$
$(4)$ says that asymptotically, the full sum from the question is $frac12sqrt{fracpi{k}}$, which vanishes as $ktoinfty$. Thus, the sum in $(18)$ must converge quickly to $fracpi2$. In fact, the sum in $(18)$ is the partial sum of the sum in this question whose limit is $fracpi2$.
answered Jan 1 at 14:22
robjohn♦robjohn
268k27308634
$endgroup$
We will make use of the identity
$$
sum_{k=0}^n(-1)^kbinom{n}{k}frac1{k+alpha}=frac{n!,Gamma(alpha)}{Gamma(n+1+alpha)}tag1
$$
which can be proven by induction on $n$.
The sum in the question is
$$
begin{align}
&sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{2n+2j+1}tag2\
&=frac12sum_{n=0}^infty(-1)^nsum_{j=0}^kbinom{k}{j}frac{(-1)^j}{n+j+frac12}tag3\
&=frac12sum_{n=0}^infty(-1)^nfrac{k!,Gamma!left(n+frac12right)}{Gamma!left(n+k+frac32right)}tag4\
&=frac12sum_{n=0}^infty(-1)^nint_0^1(1-t)^kt^{n-frac12},mathrm{d}ttag5\
&=frac12int_0^1frac{(1-t)^k}{(1+t)sqrt{t}},mathrm{d}ttag6\
&=int_0^1frac{left(1-t^2right)^k}{1+t^2},mathrm{d}ttag7\
&=int_0^1frac{left(2-left(1+t^2right)right)^k}{1+t^2},mathrm{d}ttag8\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}int_0^1left(1+t^2right)^{j-1},mathrm{d}tright)tag9\
&=2^kleft(fracpi4+sum_{j=1}^kleft(-frac12right)^jbinom{k}{j}sum_{i=0}^{j-1}binom{j-1}{i}frac1{2i+1}right)tag{10}\
&=2^kleft(fracpi4-frac12sum_{j=0}^{k-1}left(-frac12right)^jsum_{m=1}^kbinom{k-m}{j}sum_{i=0}^{j}binom{j}{i}frac1{2i+1}right)tag{11}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=i}^{k-m}left(-frac12right)^jbinom{k-m}{i}binom{k-m-i}{j-i}frac1{2i+1}right)tag{12}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}sum_{j=0}^{k-m-i}left(-frac12right)^{j+i}binom{k-m}{i}binom{k-m-i}{j}frac1{2i+1}right)tag{13}\
&=2^kleft(fracpi4-frac12sum_{m=1}^ksum_{i=0}^{k-1}left(-frac12right)^ibinom{k-m}{i}left(frac12right)^{k-m-i}frac1{2i+1}right)tag{14}\
&=2^kleft(fracpi4-frac14sum_{m=1}^ksum_{i=0}^{k-m}(-1)^ibinom{k-m}{i}left(frac12right)^{k-m}frac1{i+frac12}right)tag{15}\
&=2^kleft(fracpi4-frac14sum_{m=1}^kleft(frac12right)^{k-m}frac{(k-m)!,Gamma!left(frac12right)}{Gamma!left(k-m+frac32right)}right)tag{16}\
&=2^kleft(fracpi4-frac14sum_{m=0}^{k-1}left(frac12right)^mfrac{m!,Gamma!left(frac12right)}{Gamma!left(m+frac32right)}right)tag{17}\
&=2^kleft(fracpi4-frac12sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}right)tag{18}\
end{align}
$$
Explanation:
$phantom{1}(3)$: bring $frac12$ out front
$phantom{1}(4)$: apply $(1)$
$phantom{1}(5)$: apply the Beta Function
$phantom{1}(6)$: sum in $n$
$phantom{1}(7)$: substitute $tmapsto t^2$
$phantom{1}(8)$: $1-t^2=2-left(1+t^2right)$
$phantom{1}(9)$: apply the Binomial Theorem
$(10)$: apply the Binomial Theorem and integrate
$(11)$: $binom{k}{j}=sumlimits_{m=1}^kbinom{k-m}{j-1}$ and substitute $jmapsto j+1$
$(12)$: reorder summation and $binom{k-m}{j}binom{j}{i}=binom{k-m}{i}binom{k-m-i}{j-i}$
$(13)$: substitute $jmapsto j+i$
$(14)$: apply the Binomial Theorem
$(15)$: shift factors of $2$
$(16)$: apply $(1)$
$(17)$: substitute $mmapsto k-m$
$(18)$: rewrite the Gamma functions as factorials
Thus,
$$
bbox[5px,border:2px solid #C0A000]{g(k)=-2^{k-1}sum_{m=0}^{k-1}frac{m!}{(2m+1)!!}}tag{19}
$$
$(4)$ says that asymptotically, the full sum from the question is $frac12sqrt{fracpi{k}}$, which vanishes as $ktoinfty$. Thus, the sum in $(18)$ must converge quickly to $fracpi2$. In fact, the sum in $(18)$ is the partial sum of the sum in this question whose limit is $fracpi2$.
answered Jan 1 at 14:22
robjohn♦robjohn
268k27308634
edited Jan 2 at 2:47
answered Jan 1 at 14:22
robjohn♦robjohn
268k27308634
answered Jan 1 at 14:22
robjohn♦robjohn
268k27308634
answered Jan 1 at 14:22
robjohn♦robjohn
268k27308634
268k27308634
add a comment |
add a comment |
$begingroup$
Note that the partial fractions decomposition of the inverse
of the Rising Factorial gives:
$$
eqalign{
& left( {x - 1} right)^{,underline {, - left( {m + 1} right),} } = {1 over {x^{,overline {,m + 1,} } }}
= {{Gamma (x)} over {Gamma (x + m + 1)}} = ;quad left| {;0 le {rm integer }, m} right.quad = cr
& = sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {left( {m - k} right)!k!}}} right)}
= sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {m!}}binom{m}{k}
} right)} cr}
$$
So
$$
eqalign{
& sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {2n + 2j + 1}}} = cr
& = {1 over 2}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {left( {n + 1/2 + j} right)}}} = cr
& = {{k!} over {2left( {n + 1/2} right)^{,overline {,k + 1,} } }} cr}
$$
We have then that the sum in $n$ is
$$
eqalign{
& S(k) = sumlimits_{0, le ,n} {left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right){1 over {2n + 2j + 1}}} } = cr
& = {1 over 2}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{k!} over {left( {n + 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,k + 1,} } }
over {left( {1 + n - 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,n - 1/2,} } }
over {left( {2 + k} right)^{,overline {,n - 1/2,} } }}} = cr
& = {{1^{,overline {, - 1/2,} } } over {2left( {k + 1} right)left( {2 + k} right)^{,overline {, - 1/2,} } }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {2 + k} right)} over {2left( {k + 1} right)Gamma left( 1 right)Gamma left( {3/2 + k} right) }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {1 + k} right)} over {2,Gamma left( {3/2 + k} right)}}{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) = cr
& = 2^{,2,k} {{left( {k!} right)^{,2} } over {left( {2k + 1} right)!}};;{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) cr}
$$
where in the mid of the derivation we have used the identity
$$
{{z^{,overline {,w,} } } over {left( {z + a} right)^{,overline {,w,} } }} = {{z^{,overline {,a,} } } over {left( {z + w} right)^{,overline {,a,} } }}
$$
and at the end the duplication formula for Gamma
$$
Gamma left( {2,z} right) = {{2^{,2,z - 1} } over {sqrt pi }}Gamma left( z right)Gamma left( {z + 1/2} right) = 2^{,2,z - 1} {{Gamma left( z right)Gamma left( {z + 1/2} right)} over {Gamma left( {1/2} right)}}
$$
Alternative Approach
Actually there is a more straight alternative aproach through the Digamma Function
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {4m + 2j + 1}} - {1 over {4m + 2j + 3}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {m + j/2 + 1/4}} - {1 over {m + j/2 + 3/4}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k cr
2j cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right)left( {psi left( {j + 5/4} right) - psi left( {j + 3/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)psi left( {j + 3/4} right) - left( matrix{
k cr
2j cr} right)psi left( {j + 1/4} right) - left( matrix{
k cr
2j + 1 cr} right)psi left( {j + 1 + 1/4} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right){1 over {j + 1/4}}} cr}
$$
Since we can write (re. for instance to this page)
$$
eqalign{
& psi left( {j + 3/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 3}}} + {pi over 2} - ln 8 - gamma
= psi left( {j + 3/4} right) - left( {psi left( {3/4} right) - left( {{pi over 2} - ln 8 - gamma } right)} right) cr
& psi left( {j + 1/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 1}}} - {pi over 2} - ln 8 - gamma
= psi left( {j + 1/4} right) - left( {psi left( {1/4} right) - left( { - {pi over 2} - ln 8 - gamma } right)} right) cr}
$$
we arrive to express the sum as
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 3} right)left( {4,l + 1} right)}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 2} right)^{,2} - 1}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} cr}
$$
Interesting the comparison of this expression with the one given by
@robjohn
answered Jan 1 at 17:31
G CabG Cab
19.7k31339
$endgroup$
add a comment |
$begingroup$
Note that the partial fractions decomposition of the inverse
of the Rising Factorial gives:
$$
eqalign{
& left( {x - 1} right)^{,underline {, - left( {m + 1} right),} } = {1 over {x^{,overline {,m + 1,} } }}
= {{Gamma (x)} over {Gamma (x + m + 1)}} = ;quad left| {;0 le {rm integer }, m} right.quad = cr
& = sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {left( {m - k} right)!k!}}} right)}
= sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {m!}}binom{m}{k}
} right)} cr}
$$
So
$$
eqalign{
& sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {2n + 2j + 1}}} = cr
& = {1 over 2}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {left( {n + 1/2 + j} right)}}} = cr
& = {{k!} over {2left( {n + 1/2} right)^{,overline {,k + 1,} } }} cr}
$$
We have then that the sum in $n$ is
$$
eqalign{
& S(k) = sumlimits_{0, le ,n} {left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right){1 over {2n + 2j + 1}}} } = cr
& = {1 over 2}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{k!} over {left( {n + 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,k + 1,} } }
over {left( {1 + n - 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,n - 1/2,} } }
over {left( {2 + k} right)^{,overline {,n - 1/2,} } }}} = cr
& = {{1^{,overline {, - 1/2,} } } over {2left( {k + 1} right)left( {2 + k} right)^{,overline {, - 1/2,} } }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {2 + k} right)} over {2left( {k + 1} right)Gamma left( 1 right)Gamma left( {3/2 + k} right) }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {1 + k} right)} over {2,Gamma left( {3/2 + k} right)}}{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) = cr
& = 2^{,2,k} {{left( {k!} right)^{,2} } over {left( {2k + 1} right)!}};;{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) cr}
$$
where in the mid of the derivation we have used the identity
$$
{{z^{,overline {,w,} } } over {left( {z + a} right)^{,overline {,w,} } }} = {{z^{,overline {,a,} } } over {left( {z + w} right)^{,overline {,a,} } }}
$$
and at the end the duplication formula for Gamma
$$
Gamma left( {2,z} right) = {{2^{,2,z - 1} } over {sqrt pi }}Gamma left( z right)Gamma left( {z + 1/2} right) = 2^{,2,z - 1} {{Gamma left( z right)Gamma left( {z + 1/2} right)} over {Gamma left( {1/2} right)}}
$$
Alternative Approach
Actually there is a more straight alternative aproach through the Digamma Function
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {4m + 2j + 1}} - {1 over {4m + 2j + 3}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {m + j/2 + 1/4}} - {1 over {m + j/2 + 3/4}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k cr
2j cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right)left( {psi left( {j + 5/4} right) - psi left( {j + 3/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)psi left( {j + 3/4} right) - left( matrix{
k cr
2j cr} right)psi left( {j + 1/4} right) - left( matrix{
k cr
2j + 1 cr} right)psi left( {j + 1 + 1/4} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right){1 over {j + 1/4}}} cr}
$$
Since we can write (re. for instance to this page)
$$
eqalign{
& psi left( {j + 3/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 3}}} + {pi over 2} - ln 8 - gamma
= psi left( {j + 3/4} right) - left( {psi left( {3/4} right) - left( {{pi over 2} - ln 8 - gamma } right)} right) cr
& psi left( {j + 1/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 1}}} - {pi over 2} - ln 8 - gamma
= psi left( {j + 1/4} right) - left( {psi left( {1/4} right) - left( { - {pi over 2} - ln 8 - gamma } right)} right) cr}
$$
we arrive to express the sum as
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 3} right)left( {4,l + 1} right)}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 2} right)^{,2} - 1}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} cr}
$$
Interesting the comparison of this expression with the one given by
@robjohn
answered Jan 1 at 17:31
G CabG Cab
19.7k31339
$endgroup$
add a comment |
$begingroup$
Note that the partial fractions decomposition of the inverse
of the Rising Factorial gives:
$$
eqalign{
& left( {x - 1} right)^{,underline {, - left( {m + 1} right),} } = {1 over {x^{,overline {,m + 1,} } }}
= {{Gamma (x)} over {Gamma (x + m + 1)}} = ;quad left| {;0 le {rm integer }, m} right.quad = cr
& = sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {left( {m - k} right)!k!}}} right)}
= sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {m!}}binom{m}{k}
} right)} cr}
$$
So
$$
eqalign{
& sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {2n + 2j + 1}}} = cr
& = {1 over 2}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {left( {n + 1/2 + j} right)}}} = cr
& = {{k!} over {2left( {n + 1/2} right)^{,overline {,k + 1,} } }} cr}
$$
We have then that the sum in $n$ is
$$
eqalign{
& S(k) = sumlimits_{0, le ,n} {left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right){1 over {2n + 2j + 1}}} } = cr
& = {1 over 2}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{k!} over {left( {n + 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,k + 1,} } }
over {left( {1 + n - 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,n - 1/2,} } }
over {left( {2 + k} right)^{,overline {,n - 1/2,} } }}} = cr
& = {{1^{,overline {, - 1/2,} } } over {2left( {k + 1} right)left( {2 + k} right)^{,overline {, - 1/2,} } }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {2 + k} right)} over {2left( {k + 1} right)Gamma left( 1 right)Gamma left( {3/2 + k} right) }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {1 + k} right)} over {2,Gamma left( {3/2 + k} right)}}{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) = cr
& = 2^{,2,k} {{left( {k!} right)^{,2} } over {left( {2k + 1} right)!}};;{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) cr}
$$
where in the mid of the derivation we have used the identity
$$
{{z^{,overline {,w,} } } over {left( {z + a} right)^{,overline {,w,} } }} = {{z^{,overline {,a,} } } over {left( {z + w} right)^{,overline {,a,} } }}
$$
and at the end the duplication formula for Gamma
$$
Gamma left( {2,z} right) = {{2^{,2,z - 1} } over {sqrt pi }}Gamma left( z right)Gamma left( {z + 1/2} right) = 2^{,2,z - 1} {{Gamma left( z right)Gamma left( {z + 1/2} right)} over {Gamma left( {1/2} right)}}
$$
Alternative Approach
Actually there is a more straight alternative aproach through the Digamma Function
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {4m + 2j + 1}} - {1 over {4m + 2j + 3}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {m + j/2 + 1/4}} - {1 over {m + j/2 + 3/4}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k cr
2j cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right)left( {psi left( {j + 5/4} right) - psi left( {j + 3/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)psi left( {j + 3/4} right) - left( matrix{
k cr
2j cr} right)psi left( {j + 1/4} right) - left( matrix{
k cr
2j + 1 cr} right)psi left( {j + 1 + 1/4} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right){1 over {j + 1/4}}} cr}
$$
Since we can write (re. for instance to this page)
$$
eqalign{
& psi left( {j + 3/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 3}}} + {pi over 2} - ln 8 - gamma
= psi left( {j + 3/4} right) - left( {psi left( {3/4} right) - left( {{pi over 2} - ln 8 - gamma } right)} right) cr
& psi left( {j + 1/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 1}}} - {pi over 2} - ln 8 - gamma
= psi left( {j + 1/4} right) - left( {psi left( {1/4} right) - left( { - {pi over 2} - ln 8 - gamma } right)} right) cr}
$$
we arrive to express the sum as
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 3} right)left( {4,l + 1} right)}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 2} right)^{,2} - 1}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} cr}
$$
Interesting the comparison of this expression with the one given by
@robjohn
answered Jan 1 at 17:31
G CabG Cab
19.7k31339
$endgroup$
Note that the partial fractions decomposition of the inverse
of the Rising Factorial gives:
$$
eqalign{
& left( {x - 1} right)^{,underline {, - left( {m + 1} right),} } = {1 over {x^{,overline {,m + 1,} } }}
= {{Gamma (x)} over {Gamma (x + m + 1)}} = ;quad left| {;0 le {rm integer }, m} right.quad = cr
& = sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {left( {m - k} right)!k!}}} right)}
= sumlimits_{0, le ,k, le ,m} {{1 over {x + k}}left( {{{left( { - 1} right)^{,k} } over {m!}}binom{m}{k}
} right)} cr}
$$
So
$$
eqalign{
& sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {2n + 2j + 1}}} = cr
& = {1 over 2}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
{1 over {left( {n + 1/2 + j} right)}}} = cr
& = {{k!} over {2left( {n + 1/2} right)^{,overline {,k + 1,} } }} cr}
$$
We have then that the sum in $n$ is
$$
eqalign{
& S(k) = sumlimits_{0, le ,n} {left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right){1 over {2n + 2j + 1}}} } = cr
& = {1 over 2}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{k!} over {left( {n + 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,k + 1,} } }
over {left( {1 + n - 1/2} right)^{,overline {,k + 1,} } }}} = cr
& = {1 over {2left( {k + 1} right)}}sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{1^{,overline {,n - 1/2,} } }
over {left( {2 + k} right)^{,overline {,n - 1/2,} } }}} = cr
& = {{1^{,overline {, - 1/2,} } } over {2left( {k + 1} right)left( {2 + k} right)^{,overline {, - 1/2,} } }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {2 + k} right)} over {2left( {k + 1} right)Gamma left( 1 right)Gamma left( {3/2 + k} right) }}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {{left( {1/2} right)^{,overline {,n,} } } over {left( {3/2 + k} right)^{,overline {,n,} } }}} = cr
& = {{Gamma left( {1/2} right)Gamma left( {1 + k} right)} over {2,Gamma left( {3/2 + k} right)}}{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) = cr
& = 2^{,2,k} {{left( {k!} right)^{,2} } over {left( {2k + 1} right)!}};;{}_2F_{,1} left( {left. {matrix{
{1/2;,;1} cr
{k + 3/2} cr
} ;} right| - 1} right) cr}
$$
where in the mid of the derivation we have used the identity
$$
{{z^{,overline {,w,} } } over {left( {z + a} right)^{,overline {,w,} } }} = {{z^{,overline {,a,} } } over {left( {z + w} right)^{,overline {,a,} } }}
$$
and at the end the duplication formula for Gamma
$$
Gamma left( {2,z} right) = {{2^{,2,z - 1} } over {sqrt pi }}Gamma left( z right)Gamma left( {z + 1/2} right) = 2^{,2,z - 1} {{Gamma left( z right)Gamma left( {z + 1/2} right)} over {Gamma left( {1/2} right)}}
$$
Alternative Approach
Actually there is a more straight alternative aproach through the Digamma Function
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {4m + 2j + 1}} - {1 over {4m + 2j + 3}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)sumlimits_{0, le ,m} {left( {{1 over {m + j/2 + 1/4}} - {1 over {m + j/2 + 3/4}}} right)} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} left( matrix{
k cr
j cr} right)left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k cr
2j cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right)left( {psi left( {j + 5/4} right) - psi left( {j + 3/4} right)} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)psi left( {j + 3/4} right) - left( matrix{
k cr
2j cr} right)psi left( {j + 1/4} right) - left( matrix{
k cr
2j + 1 cr} right)psi left( {j + 1 + 1/4} right)} = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,} {left( matrix{
k + 1 cr
2j + 1 cr} right)left( {psi left( {j + 3/4} right) - psi left( {j + 1/4} right)} right) - left( matrix{
k cr
2j + 1 cr} right){1 over {j + 1/4}}} cr}
$$
Since we can write (re. for instance to this page)
$$
eqalign{
& psi left( {j + 3/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 3}}} + {pi over 2} - ln 8 - gamma
= psi left( {j + 3/4} right) - left( {psi left( {3/4} right) - left( {{pi over 2} - ln 8 - gamma } right)} right) cr
& psi left( {j + 1/4} right) = 4sumlimits_{0, le ,l, le ,j - 1,} {{1 over {4,l + 1}}} - {pi over 2} - ln 8 - gamma
= psi left( {j + 1/4} right) - left( {psi left( {1/4} right) - left( { - {pi over 2} - ln 8 - gamma } right)} right) cr}
$$
we arrive to express the sum as
$$
eqalign{
& S(k) = sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
sumlimits_{0, le ,n} {left( { - 1} right)^{,n} {1 over {2n + 2j + 1}}} } = cr
& = {1 over 4}sumlimits_{left( {0, le } right),j,left( { le ,k} right)} {left( { - 1} right)^{,j} binom{k}{j}
left( {psi left( {j/2 + 3/4} right) - psi left( {j/2 + 1/4} right)} right)} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 3} right)left( {4,l + 1} right)}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} = cr
& = {pi over 4}2^{,k} - 2sumlimits_{left( {0, le } right),j,left( { le ,k} right),} {binom{k+1}{2j+1}
left( {sumlimits_{0, le ,l, le ,j - 1,} {{1 over {left( {4,l + 2} right)^{,2} - 1}}} } right)}
- sumlimits_{left( {0, le } right),j,left( { le ,left( {k - 1} right)/2} right),} {binom{k}{2j+1}{1 over {4j + 1}}} cr}
$$
Interesting the comparison of this expression with the one given by
@robjohn
answered Jan 1 at 17:31
G CabG Cab
19.7k31339
edited Jan 3 at 11:57
answered Jan 1 at 17:31
G CabG Cab
19.7k31339
answered Jan 1 at 17:31
G CabG Cab
19.7k31339
answered Jan 1 at 17:31
G CabG Cab
19.7k31339
19.7k31339
add a comment |
add a comment |
$begingroup$
(2) is
$$frac11-frac23+frac25-frac27+cdots=-1+2left(frac11-frac13+frac15-cdotsright),$$
(3) is
$$frac11-frac33+frac45-frac47+cdots=-3+frac13+4left(frac11-frac13+frac15-cdotsright),$$
(4) is
$$frac11-frac43+frac75-frac87+cdots=-7+frac43-frac15+8left(frac11-frac13+frac15-cdotsright),$$
etc.
In general, one gets
$$2^kfracpi4-sum_{j=0}^{k-1}(-1)^jfrac1{2j+1}
left(2^k-sum_{i=0}^j{kchoose i}right).$$
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
add a comment |
$begingroup$
(2) is
$$frac11-frac23+frac25-frac27+cdots=-1+2left(frac11-frac13+frac15-cdotsright),$$
(3) is
$$frac11-frac33+frac45-frac47+cdots=-3+frac13+4left(frac11-frac13+frac15-cdotsright),$$
(4) is
$$frac11-frac43+frac75-frac87+cdots=-7+frac43-frac15+8left(frac11-frac13+frac15-cdotsright),$$
etc.
In general, one gets
$$2^kfracpi4-sum_{j=0}^{k-1}(-1)^jfrac1{2j+1}
left(2^k-sum_{i=0}^j{kchoose i}right).$$
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
add a comment |
$begingroup$
(2) is
$$frac11-frac23+frac25-frac27+cdots=-1+2left(frac11-frac13+frac15-cdotsright),$$
(3) is
$$frac11-frac33+frac45-frac47+cdots=-3+frac13+4left(frac11-frac13+frac15-cdotsright),$$
(4) is
$$frac11-frac43+frac75-frac87+cdots=-7+frac43-frac15+8left(frac11-frac13+frac15-cdotsright),$$
etc.
In general, one gets
$$2^kfracpi4-sum_{j=0}^{k-1}(-1)^jfrac1{2j+1}
left(2^k-sum_{i=0}^j{kchoose i}right).$$
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
(2) is
$$frac11-frac23+frac25-frac27+cdots=-1+2left(frac11-frac13+frac15-cdotsright),$$
(3) is
$$frac11-frac33+frac45-frac47+cdots=-3+frac13+4left(frac11-frac13+frac15-cdotsright),$$
(4) is
$$frac11-frac43+frac75-frac87+cdots=-7+frac43-frac15+8left(frac11-frac13+frac15-cdotsright),$$
etc.
In general, one gets
$$2^kfracpi4-sum_{j=0}^{k-1}(-1)^jfrac1{2j+1}
left(2^k-sum_{i=0}^j{kchoose i}right).$$
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
105k1160133
edited Dec 31 '18 at 16:44
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 31 '18 at 16:38
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
add a comment |
add a comment |
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