Prove:...












16












$begingroup$


Consider the following limit:
$$lim_{ntoinfty}{sum_{m=0}^{n}{sum_{k=0}^{n-m}{left[frac{2^{n-m-k}}{n-m+1},frac{{{2k}choose{k}}{{2m}choose{m}}}{{{2n}choose{n}}}right]}}}=pi$$
I cooked this up while playing around with power series (details below).



Is there a more direct way to prove this limit?





Consider the functions $f(x)=frac{tan^{-1}(sqrt{1-x})}{sqrt{1-x}}$, $g(x)=frac{pi/2-tan^{-1}(sqrt{1-x})}{sqrt{1-x}}$. We write the power series:
$$f(x)=sum_{n=0}^{infty}{(s_npi-r_n)x^n},qquad{g}(x)=sum_{n=0}^{infty}{(s_npi+r_n)x^n}$$
where computing the first few terms suggests that $r_n$, $s_n$ are rational.



Indeed, we have $frac{pi/4-tan^{-1}(sqrt{1-x})}{sqrt{1-x}}=frac{g(x)-f(x)}{2}=sum{r_nx^n}$, and $frac{1/4}{sqrt{1-x}}=frac{g(x)+f(x)}{2pi}=sum{s_nx^n}$.



Using $frac{d}{dx}[tan^{-1}(sqrt{1-x})]=-frac{1}{2(2-x)sqrt{1-x}}$, we can use the power series for $frac{1}{2-x}$ and $frac{1}{sqrt{1-x}}$ to calculate the power series for $(pi/4-tan^{-1}(sqrt{1-x}))$, which consists of only rational coefficients. Combining this with the power series for $frac{1}{sqrt{1-x}}$ gives:
$$r_n=frac{1}{2}sum_{m=0}^{n-1}{sum_{k=0}^{n-m-1}{frac{{{2k}choose{k}}{{2m}choose{m}}}{(n-m)cdot2^{k+m+n}}}}$$
We easily get $s_n=frac{1}{2^{2n+2}}{2nchoose{n}}$.



Now, because no branch of $f(z)$ has singularities anywhere (the apparent singularity at $z=1$ is removable), the coefficients of the power series of $f$ must tend to zero.



Hence $lim_{ntoinfty}{frac{r_{n+1}}{s_{n+1}}}=pi$, and the desired limit follows after simplifying.





Notes:



1) It is easily shown that:
$$s_npi-r_n=int_{0}^{pi/4}{sin^{2n}{theta},dtheta},qquad{s}_npi+r_n=int_{pi/4}^{pi/2}{sin^{2n}{theta},dtheta}$$



2) The above proof actually shows:
$$lim_{ntoinfty}{left(1+frac{1}{2n+1}right)sum_{m=0}^{n}{sum_{k=0}^{n-m}{left[frac{2^{n-m-k}}{n-m+1},frac{{{2k}choose{k}}{{2m}choose{m}}}{{{2n}choose{n}}}right]}}}=pi$$
which converges much faster than the given limit.










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Holy smokes....
    $endgroup$
    – Barron
    Dec 31 '18 at 19:01
















16












$begingroup$


Consider the following limit:
$$lim_{ntoinfty}{sum_{m=0}^{n}{sum_{k=0}^{n-m}{left[frac{2^{n-m-k}}{n-m+1},frac{{{2k}choose{k}}{{2m}choose{m}}}{{{2n}choose{n}}}right]}}}=pi$$
I cooked this up while playing around with power series (details below).



Is there a more direct way to prove this limit?





Consider the functions $f(x)=frac{tan^{-1}(sqrt{1-x})}{sqrt{1-x}}$, $g(x)=frac{pi/2-tan^{-1}(sqrt{1-x})}{sqrt{1-x}}$. We write the power series:
$$f(x)=sum_{n=0}^{infty}{(s_npi-r_n)x^n},qquad{g}(x)=sum_{n=0}^{infty}{(s_npi+r_n)x^n}$$
where computing the first few terms suggests that $r_n$, $s_n$ are rational.



Indeed, we have $frac{pi/4-tan^{-1}(sqrt{1-x})}{sqrt{1-x}}=frac{g(x)-f(x)}{2}=sum{r_nx^n}$, and $frac{1/4}{sqrt{1-x}}=frac{g(x)+f(x)}{2pi}=sum{s_nx^n}$.



Using $frac{d}{dx}[tan^{-1}(sqrt{1-x})]=-frac{1}{2(2-x)sqrt{1-x}}$, we can use the power series for $frac{1}{2-x}$ and $frac{1}{sqrt{1-x}}$ to calculate the power series for $(pi/4-tan^{-1}(sqrt{1-x}))$, which consists of only rational coefficients. Combining this with the power series for $frac{1}{sqrt{1-x}}$ gives:
$$r_n=frac{1}{2}sum_{m=0}^{n-1}{sum_{k=0}^{n-m-1}{frac{{{2k}choose{k}}{{2m}choose{m}}}{(n-m)cdot2^{k+m+n}}}}$$
We easily get $s_n=frac{1}{2^{2n+2}}{2nchoose{n}}$.



Now, because no branch of $f(z)$ has singularities anywhere (the apparent singularity at $z=1$ is removable), the coefficients of the power series of $f$ must tend to zero.



Hence $lim_{ntoinfty}{frac{r_{n+1}}{s_{n+1}}}=pi$, and the desired limit follows after simplifying.





Notes:



1) It is easily shown that:
$$s_npi-r_n=int_{0}^{pi/4}{sin^{2n}{theta},dtheta},qquad{s}_npi+r_n=int_{pi/4}^{pi/2}{sin^{2n}{theta},dtheta}$$



2) The above proof actually shows:
$$lim_{ntoinfty}{left(1+frac{1}{2n+1}right)sum_{m=0}^{n}{sum_{k=0}^{n-m}{left[frac{2^{n-m-k}}{n-m+1},frac{{{2k}choose{k}}{{2m}choose{m}}}{{{2n}choose{n}}}right]}}}=pi$$
which converges much faster than the given limit.










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Holy smokes....
    $endgroup$
    – Barron
    Dec 31 '18 at 19:01














16












16








16


4



$begingroup$


Consider the following limit:
$$lim_{ntoinfty}{sum_{m=0}^{n}{sum_{k=0}^{n-m}{left[frac{2^{n-m-k}}{n-m+1},frac{{{2k}choose{k}}{{2m}choose{m}}}{{{2n}choose{n}}}right]}}}=pi$$
I cooked this up while playing around with power series (details below).



Is there a more direct way to prove this limit?





Consider the functions $f(x)=frac{tan^{-1}(sqrt{1-x})}{sqrt{1-x}}$, $g(x)=frac{pi/2-tan^{-1}(sqrt{1-x})}{sqrt{1-x}}$. We write the power series:
$$f(x)=sum_{n=0}^{infty}{(s_npi-r_n)x^n},qquad{g}(x)=sum_{n=0}^{infty}{(s_npi+r_n)x^n}$$
where computing the first few terms suggests that $r_n$, $s_n$ are rational.



Indeed, we have $frac{pi/4-tan^{-1}(sqrt{1-x})}{sqrt{1-x}}=frac{g(x)-f(x)}{2}=sum{r_nx^n}$, and $frac{1/4}{sqrt{1-x}}=frac{g(x)+f(x)}{2pi}=sum{s_nx^n}$.



Using $frac{d}{dx}[tan^{-1}(sqrt{1-x})]=-frac{1}{2(2-x)sqrt{1-x}}$, we can use the power series for $frac{1}{2-x}$ and $frac{1}{sqrt{1-x}}$ to calculate the power series for $(pi/4-tan^{-1}(sqrt{1-x}))$, which consists of only rational coefficients. Combining this with the power series for $frac{1}{sqrt{1-x}}$ gives:
$$r_n=frac{1}{2}sum_{m=0}^{n-1}{sum_{k=0}^{n-m-1}{frac{{{2k}choose{k}}{{2m}choose{m}}}{(n-m)cdot2^{k+m+n}}}}$$
We easily get $s_n=frac{1}{2^{2n+2}}{2nchoose{n}}$.



Now, because no branch of $f(z)$ has singularities anywhere (the apparent singularity at $z=1$ is removable), the coefficients of the power series of $f$ must tend to zero.



Hence $lim_{ntoinfty}{frac{r_{n+1}}{s_{n+1}}}=pi$, and the desired limit follows after simplifying.





Notes:



1) It is easily shown that:
$$s_npi-r_n=int_{0}^{pi/4}{sin^{2n}{theta},dtheta},qquad{s}_npi+r_n=int_{pi/4}^{pi/2}{sin^{2n}{theta},dtheta}$$



2) The above proof actually shows:
$$lim_{ntoinfty}{left(1+frac{1}{2n+1}right)sum_{m=0}^{n}{sum_{k=0}^{n-m}{left[frac{2^{n-m-k}}{n-m+1},frac{{{2k}choose{k}}{{2m}choose{m}}}{{{2n}choose{n}}}right]}}}=pi$$
which converges much faster than the given limit.










share|cite|improve this question











$endgroup$




Consider the following limit:
$$lim_{ntoinfty}{sum_{m=0}^{n}{sum_{k=0}^{n-m}{left[frac{2^{n-m-k}}{n-m+1},frac{{{2k}choose{k}}{{2m}choose{m}}}{{{2n}choose{n}}}right]}}}=pi$$
I cooked this up while playing around with power series (details below).



Is there a more direct way to prove this limit?





Consider the functions $f(x)=frac{tan^{-1}(sqrt{1-x})}{sqrt{1-x}}$, $g(x)=frac{pi/2-tan^{-1}(sqrt{1-x})}{sqrt{1-x}}$. We write the power series:
$$f(x)=sum_{n=0}^{infty}{(s_npi-r_n)x^n},qquad{g}(x)=sum_{n=0}^{infty}{(s_npi+r_n)x^n}$$
where computing the first few terms suggests that $r_n$, $s_n$ are rational.



Indeed, we have $frac{pi/4-tan^{-1}(sqrt{1-x})}{sqrt{1-x}}=frac{g(x)-f(x)}{2}=sum{r_nx^n}$, and $frac{1/4}{sqrt{1-x}}=frac{g(x)+f(x)}{2pi}=sum{s_nx^n}$.



Using $frac{d}{dx}[tan^{-1}(sqrt{1-x})]=-frac{1}{2(2-x)sqrt{1-x}}$, we can use the power series for $frac{1}{2-x}$ and $frac{1}{sqrt{1-x}}$ to calculate the power series for $(pi/4-tan^{-1}(sqrt{1-x}))$, which consists of only rational coefficients. Combining this with the power series for $frac{1}{sqrt{1-x}}$ gives:
$$r_n=frac{1}{2}sum_{m=0}^{n-1}{sum_{k=0}^{n-m-1}{frac{{{2k}choose{k}}{{2m}choose{m}}}{(n-m)cdot2^{k+m+n}}}}$$
We easily get $s_n=frac{1}{2^{2n+2}}{2nchoose{n}}$.



Now, because no branch of $f(z)$ has singularities anywhere (the apparent singularity at $z=1$ is removable), the coefficients of the power series of $f$ must tend to zero.



Hence $lim_{ntoinfty}{frac{r_{n+1}}{s_{n+1}}}=pi$, and the desired limit follows after simplifying.





Notes:



1) It is easily shown that:
$$s_npi-r_n=int_{0}^{pi/4}{sin^{2n}{theta},dtheta},qquad{s}_npi+r_n=int_{pi/4}^{pi/2}{sin^{2n}{theta},dtheta}$$



2) The above proof actually shows:
$$lim_{ntoinfty}{left(1+frac{1}{2n+1}right)sum_{m=0}^{n}{sum_{k=0}^{n-m}{left[frac{2^{n-m-k}}{n-m+1},frac{{{2k}choose{k}}{{2m}choose{m}}}{{{2n}choose{n}}}right]}}}=pi$$
which converges much faster than the given limit.







limits proof-verification summation binomial-coefficients alternative-proof






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 23:30







user630758

















asked Dec 31 '18 at 18:54









AntAnt

1,645826




1,645826








  • 3




    $begingroup$
    Holy smokes....
    $endgroup$
    – Barron
    Dec 31 '18 at 19:01














  • 3




    $begingroup$
    Holy smokes....
    $endgroup$
    – Barron
    Dec 31 '18 at 19:01








3




3




$begingroup$
Holy smokes....
$endgroup$
– Barron
Dec 31 '18 at 19:01




$begingroup$
Holy smokes....
$endgroup$
– Barron
Dec 31 '18 at 19:01










1 Answer
1






active

oldest

votes


















4












$begingroup$


I think OP did a really good job and this answer aims to indicate that it is plausible to obtain the specific type of generating functions like $arctan$ as stated by OP. Here we start with the binomial expression
begin{align*}
q_n:=sum_{m=0}^nsum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}frac{2^{n-m-k}}{n-m+1}tag{1}
end{align*}

which corrresponds to OPs limit expression without the factor $binom{2n}{n}^{-1}$ and derive from it a generating function.



Note that since $q_n=frac{1}{4^n}r_{n+1}$ OPs claim can be stated as
begin{align*}
q_nsim pibinom{2n}{n}simsqrt{frac{pi}{n}}cdot 4^n
end{align*}

where we use the asymptotic formula of the central binomial coefficient.




Two aspects:




  • We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. Recalling the generating function of the central binomial coefficient we can write for instance
    begin{align*}
    [z^n]frac{1}{sqrt{1-4z}}=binom{2n}{n}tag{2}
    end{align*}


  • We can sum up coefficients $a_n$ by multiplication with $frac{1}{1-z}$. If $A(z)=sum_{n=0}^infty a_nz^n$ we have
    begin{align*}
    frac{1}{1-z}A(z)&=sum_{n=0}^inftyleft( sum_{k=0}^na_kright)z^n
    end{align*}

    Somewhat more general by multiplication with $frac{1}{1-pz}$ we have
    begin{align*}
    frac{1}{1-pz}A(z)&=sum_{n=0}^infty left(sum_{k=0}^na_kp^{n-k}right) z^ntag{3}
    end{align*}




We obtain
begin{align*}
color{blue}{sum_{m=0}^n}&color{blue}{sum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}frac{2^{n-m-k}}{n-m+1}}\
&=int_{0}^1sum_{m=0}^nsum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}2^{n-m-k}z^{n-m},dztag{4}\
&=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}sum_{k=0}^{n-m}binom{2k}{k}2^{n-m-k},dztag{5}\
&=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}[t^{n-m}]frac{1}{(1-2t)sqrt{1-4t}},dztag{6}\
&=int_{0}^1sum_{m=0}^inftybinom{2n-2m}{n-m}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{7}\
&=int_{0}^inftysum_{m=0}^infty[u^{n-m}]frac{1}{sqrt{1-4u}}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{8}\
&=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1sum_{m=0}^infty(zu)^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{9}\
&=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1frac{1}{(1-2zu)sqrt{1-4zu}},dztag{10}\
&=[u^n]frac{1}{sqrt{1-4u}}left.frac{arctanleft(sqrt{1-4zu}right)}{u}right|_{z=0}^{z=1}tag{11}\
&,,color{blue}{=[u^{n+1}]frac{1}{sqrt{1-4u}}left(-arctanleft(sqrt{1-4u}right)+frac{pi}{4}right)}tag{12}
end{align*}

and we also get when deriving generating functions directly from (1) the same function as OP. The scaling factor $4$ in $sqrt{1-4u}$ is from formula (2) and indicates the connection between $q_n$ and $r_n$ as stated at the beginning of this post.




Comment:




  • In (4) we use $frac{1}{p+1}=int_0^1z^{p},dz$ where $pne -1$.


  • In (5) we do a rearrangement only.


  • In (6) we apply the coefficient of operator by using (2) and (3) with $p=2$.


  • In (7) we change the order of summation by $mto n-m$ and we replace the upper index $n$ by $infty$ without changing anything, since $binom{2n-2m}{n-m}=0$ when $m>n$.


  • In (8) we apply again the coefficient of operator to $binom{2n-2m}{n-m}$ according to (2).


  • In (9) we use the linearity of the operators and apply the rule $[u^{p-q}]A(u)=[u^p]u^qA(u)$.


  • In (10) we apply the substitution rule of the coefficient of operator with $t=zu$
    begin{align*}
    A(z)=sum_{m=0}^infty a_m z^m=sum_{m=0}^infty z^m [u^m]A(u)
    end{align*}


  • In (11) we integrate obtaining the $arctan$ function.


  • In (12) we finally evaluate the $arctan$ function at lower and upper limit and apply again the rule $[u^n]frac{1}{u}A(u)=[u^{n+1}]A(u)$ as we did in (10).







share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057956%2fprove-lim-n-to-infty-sum-m-0n-sum-k-0n-m-frac2n-m-kn-m1%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$


    I think OP did a really good job and this answer aims to indicate that it is plausible to obtain the specific type of generating functions like $arctan$ as stated by OP. Here we start with the binomial expression
    begin{align*}
    q_n:=sum_{m=0}^nsum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}frac{2^{n-m-k}}{n-m+1}tag{1}
    end{align*}

    which corrresponds to OPs limit expression without the factor $binom{2n}{n}^{-1}$ and derive from it a generating function.



    Note that since $q_n=frac{1}{4^n}r_{n+1}$ OPs claim can be stated as
    begin{align*}
    q_nsim pibinom{2n}{n}simsqrt{frac{pi}{n}}cdot 4^n
    end{align*}

    where we use the asymptotic formula of the central binomial coefficient.




    Two aspects:




    • We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. Recalling the generating function of the central binomial coefficient we can write for instance
      begin{align*}
      [z^n]frac{1}{sqrt{1-4z}}=binom{2n}{n}tag{2}
      end{align*}


    • We can sum up coefficients $a_n$ by multiplication with $frac{1}{1-z}$. If $A(z)=sum_{n=0}^infty a_nz^n$ we have
      begin{align*}
      frac{1}{1-z}A(z)&=sum_{n=0}^inftyleft( sum_{k=0}^na_kright)z^n
      end{align*}

      Somewhat more general by multiplication with $frac{1}{1-pz}$ we have
      begin{align*}
      frac{1}{1-pz}A(z)&=sum_{n=0}^infty left(sum_{k=0}^na_kp^{n-k}right) z^ntag{3}
      end{align*}




    We obtain
    begin{align*}
    color{blue}{sum_{m=0}^n}&color{blue}{sum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}frac{2^{n-m-k}}{n-m+1}}\
    &=int_{0}^1sum_{m=0}^nsum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}2^{n-m-k}z^{n-m},dztag{4}\
    &=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}sum_{k=0}^{n-m}binom{2k}{k}2^{n-m-k},dztag{5}\
    &=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}[t^{n-m}]frac{1}{(1-2t)sqrt{1-4t}},dztag{6}\
    &=int_{0}^1sum_{m=0}^inftybinom{2n-2m}{n-m}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{7}\
    &=int_{0}^inftysum_{m=0}^infty[u^{n-m}]frac{1}{sqrt{1-4u}}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{8}\
    &=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1sum_{m=0}^infty(zu)^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{9}\
    &=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1frac{1}{(1-2zu)sqrt{1-4zu}},dztag{10}\
    &=[u^n]frac{1}{sqrt{1-4u}}left.frac{arctanleft(sqrt{1-4zu}right)}{u}right|_{z=0}^{z=1}tag{11}\
    &,,color{blue}{=[u^{n+1}]frac{1}{sqrt{1-4u}}left(-arctanleft(sqrt{1-4u}right)+frac{pi}{4}right)}tag{12}
    end{align*}

    and we also get when deriving generating functions directly from (1) the same function as OP. The scaling factor $4$ in $sqrt{1-4u}$ is from formula (2) and indicates the connection between $q_n$ and $r_n$ as stated at the beginning of this post.




    Comment:




    • In (4) we use $frac{1}{p+1}=int_0^1z^{p},dz$ where $pne -1$.


    • In (5) we do a rearrangement only.


    • In (6) we apply the coefficient of operator by using (2) and (3) with $p=2$.


    • In (7) we change the order of summation by $mto n-m$ and we replace the upper index $n$ by $infty$ without changing anything, since $binom{2n-2m}{n-m}=0$ when $m>n$.


    • In (8) we apply again the coefficient of operator to $binom{2n-2m}{n-m}$ according to (2).


    • In (9) we use the linearity of the operators and apply the rule $[u^{p-q}]A(u)=[u^p]u^qA(u)$.


    • In (10) we apply the substitution rule of the coefficient of operator with $t=zu$
      begin{align*}
      A(z)=sum_{m=0}^infty a_m z^m=sum_{m=0}^infty z^m [u^m]A(u)
      end{align*}


    • In (11) we integrate obtaining the $arctan$ function.


    • In (12) we finally evaluate the $arctan$ function at lower and upper limit and apply again the rule $[u^n]frac{1}{u}A(u)=[u^{n+1}]A(u)$ as we did in (10).







    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$


      I think OP did a really good job and this answer aims to indicate that it is plausible to obtain the specific type of generating functions like $arctan$ as stated by OP. Here we start with the binomial expression
      begin{align*}
      q_n:=sum_{m=0}^nsum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}frac{2^{n-m-k}}{n-m+1}tag{1}
      end{align*}

      which corrresponds to OPs limit expression without the factor $binom{2n}{n}^{-1}$ and derive from it a generating function.



      Note that since $q_n=frac{1}{4^n}r_{n+1}$ OPs claim can be stated as
      begin{align*}
      q_nsim pibinom{2n}{n}simsqrt{frac{pi}{n}}cdot 4^n
      end{align*}

      where we use the asymptotic formula of the central binomial coefficient.




      Two aspects:




      • We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. Recalling the generating function of the central binomial coefficient we can write for instance
        begin{align*}
        [z^n]frac{1}{sqrt{1-4z}}=binom{2n}{n}tag{2}
        end{align*}


      • We can sum up coefficients $a_n$ by multiplication with $frac{1}{1-z}$. If $A(z)=sum_{n=0}^infty a_nz^n$ we have
        begin{align*}
        frac{1}{1-z}A(z)&=sum_{n=0}^inftyleft( sum_{k=0}^na_kright)z^n
        end{align*}

        Somewhat more general by multiplication with $frac{1}{1-pz}$ we have
        begin{align*}
        frac{1}{1-pz}A(z)&=sum_{n=0}^infty left(sum_{k=0}^na_kp^{n-k}right) z^ntag{3}
        end{align*}




      We obtain
      begin{align*}
      color{blue}{sum_{m=0}^n}&color{blue}{sum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}frac{2^{n-m-k}}{n-m+1}}\
      &=int_{0}^1sum_{m=0}^nsum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}2^{n-m-k}z^{n-m},dztag{4}\
      &=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}sum_{k=0}^{n-m}binom{2k}{k}2^{n-m-k},dztag{5}\
      &=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}[t^{n-m}]frac{1}{(1-2t)sqrt{1-4t}},dztag{6}\
      &=int_{0}^1sum_{m=0}^inftybinom{2n-2m}{n-m}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{7}\
      &=int_{0}^inftysum_{m=0}^infty[u^{n-m}]frac{1}{sqrt{1-4u}}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{8}\
      &=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1sum_{m=0}^infty(zu)^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{9}\
      &=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1frac{1}{(1-2zu)sqrt{1-4zu}},dztag{10}\
      &=[u^n]frac{1}{sqrt{1-4u}}left.frac{arctanleft(sqrt{1-4zu}right)}{u}right|_{z=0}^{z=1}tag{11}\
      &,,color{blue}{=[u^{n+1}]frac{1}{sqrt{1-4u}}left(-arctanleft(sqrt{1-4u}right)+frac{pi}{4}right)}tag{12}
      end{align*}

      and we also get when deriving generating functions directly from (1) the same function as OP. The scaling factor $4$ in $sqrt{1-4u}$ is from formula (2) and indicates the connection between $q_n$ and $r_n$ as stated at the beginning of this post.




      Comment:




      • In (4) we use $frac{1}{p+1}=int_0^1z^{p},dz$ where $pne -1$.


      • In (5) we do a rearrangement only.


      • In (6) we apply the coefficient of operator by using (2) and (3) with $p=2$.


      • In (7) we change the order of summation by $mto n-m$ and we replace the upper index $n$ by $infty$ without changing anything, since $binom{2n-2m}{n-m}=0$ when $m>n$.


      • In (8) we apply again the coefficient of operator to $binom{2n-2m}{n-m}$ according to (2).


      • In (9) we use the linearity of the operators and apply the rule $[u^{p-q}]A(u)=[u^p]u^qA(u)$.


      • In (10) we apply the substitution rule of the coefficient of operator with $t=zu$
        begin{align*}
        A(z)=sum_{m=0}^infty a_m z^m=sum_{m=0}^infty z^m [u^m]A(u)
        end{align*}


      • In (11) we integrate obtaining the $arctan$ function.


      • In (12) we finally evaluate the $arctan$ function at lower and upper limit and apply again the rule $[u^n]frac{1}{u}A(u)=[u^{n+1}]A(u)$ as we did in (10).







      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$


        I think OP did a really good job and this answer aims to indicate that it is plausible to obtain the specific type of generating functions like $arctan$ as stated by OP. Here we start with the binomial expression
        begin{align*}
        q_n:=sum_{m=0}^nsum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}frac{2^{n-m-k}}{n-m+1}tag{1}
        end{align*}

        which corrresponds to OPs limit expression without the factor $binom{2n}{n}^{-1}$ and derive from it a generating function.



        Note that since $q_n=frac{1}{4^n}r_{n+1}$ OPs claim can be stated as
        begin{align*}
        q_nsim pibinom{2n}{n}simsqrt{frac{pi}{n}}cdot 4^n
        end{align*}

        where we use the asymptotic formula of the central binomial coefficient.




        Two aspects:




        • We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. Recalling the generating function of the central binomial coefficient we can write for instance
          begin{align*}
          [z^n]frac{1}{sqrt{1-4z}}=binom{2n}{n}tag{2}
          end{align*}


        • We can sum up coefficients $a_n$ by multiplication with $frac{1}{1-z}$. If $A(z)=sum_{n=0}^infty a_nz^n$ we have
          begin{align*}
          frac{1}{1-z}A(z)&=sum_{n=0}^inftyleft( sum_{k=0}^na_kright)z^n
          end{align*}

          Somewhat more general by multiplication with $frac{1}{1-pz}$ we have
          begin{align*}
          frac{1}{1-pz}A(z)&=sum_{n=0}^infty left(sum_{k=0}^na_kp^{n-k}right) z^ntag{3}
          end{align*}




        We obtain
        begin{align*}
        color{blue}{sum_{m=0}^n}&color{blue}{sum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}frac{2^{n-m-k}}{n-m+1}}\
        &=int_{0}^1sum_{m=0}^nsum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}2^{n-m-k}z^{n-m},dztag{4}\
        &=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}sum_{k=0}^{n-m}binom{2k}{k}2^{n-m-k},dztag{5}\
        &=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}[t^{n-m}]frac{1}{(1-2t)sqrt{1-4t}},dztag{6}\
        &=int_{0}^1sum_{m=0}^inftybinom{2n-2m}{n-m}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{7}\
        &=int_{0}^inftysum_{m=0}^infty[u^{n-m}]frac{1}{sqrt{1-4u}}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{8}\
        &=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1sum_{m=0}^infty(zu)^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{9}\
        &=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1frac{1}{(1-2zu)sqrt{1-4zu}},dztag{10}\
        &=[u^n]frac{1}{sqrt{1-4u}}left.frac{arctanleft(sqrt{1-4zu}right)}{u}right|_{z=0}^{z=1}tag{11}\
        &,,color{blue}{=[u^{n+1}]frac{1}{sqrt{1-4u}}left(-arctanleft(sqrt{1-4u}right)+frac{pi}{4}right)}tag{12}
        end{align*}

        and we also get when deriving generating functions directly from (1) the same function as OP. The scaling factor $4$ in $sqrt{1-4u}$ is from formula (2) and indicates the connection between $q_n$ and $r_n$ as stated at the beginning of this post.




        Comment:




        • In (4) we use $frac{1}{p+1}=int_0^1z^{p},dz$ where $pne -1$.


        • In (5) we do a rearrangement only.


        • In (6) we apply the coefficient of operator by using (2) and (3) with $p=2$.


        • In (7) we change the order of summation by $mto n-m$ and we replace the upper index $n$ by $infty$ without changing anything, since $binom{2n-2m}{n-m}=0$ when $m>n$.


        • In (8) we apply again the coefficient of operator to $binom{2n-2m}{n-m}$ according to (2).


        • In (9) we use the linearity of the operators and apply the rule $[u^{p-q}]A(u)=[u^p]u^qA(u)$.


        • In (10) we apply the substitution rule of the coefficient of operator with $t=zu$
          begin{align*}
          A(z)=sum_{m=0}^infty a_m z^m=sum_{m=0}^infty z^m [u^m]A(u)
          end{align*}


        • In (11) we integrate obtaining the $arctan$ function.


        • In (12) we finally evaluate the $arctan$ function at lower and upper limit and apply again the rule $[u^n]frac{1}{u}A(u)=[u^{n+1}]A(u)$ as we did in (10).







        share|cite|improve this answer











        $endgroup$




        I think OP did a really good job and this answer aims to indicate that it is plausible to obtain the specific type of generating functions like $arctan$ as stated by OP. Here we start with the binomial expression
        begin{align*}
        q_n:=sum_{m=0}^nsum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}frac{2^{n-m-k}}{n-m+1}tag{1}
        end{align*}

        which corrresponds to OPs limit expression without the factor $binom{2n}{n}^{-1}$ and derive from it a generating function.



        Note that since $q_n=frac{1}{4^n}r_{n+1}$ OPs claim can be stated as
        begin{align*}
        q_nsim pibinom{2n}{n}simsqrt{frac{pi}{n}}cdot 4^n
        end{align*}

        where we use the asymptotic formula of the central binomial coefficient.




        Two aspects:




        • We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. Recalling the generating function of the central binomial coefficient we can write for instance
          begin{align*}
          [z^n]frac{1}{sqrt{1-4z}}=binom{2n}{n}tag{2}
          end{align*}


        • We can sum up coefficients $a_n$ by multiplication with $frac{1}{1-z}$. If $A(z)=sum_{n=0}^infty a_nz^n$ we have
          begin{align*}
          frac{1}{1-z}A(z)&=sum_{n=0}^inftyleft( sum_{k=0}^na_kright)z^n
          end{align*}

          Somewhat more general by multiplication with $frac{1}{1-pz}$ we have
          begin{align*}
          frac{1}{1-pz}A(z)&=sum_{n=0}^infty left(sum_{k=0}^na_kp^{n-k}right) z^ntag{3}
          end{align*}




        We obtain
        begin{align*}
        color{blue}{sum_{m=0}^n}&color{blue}{sum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}frac{2^{n-m-k}}{n-m+1}}\
        &=int_{0}^1sum_{m=0}^nsum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}2^{n-m-k}z^{n-m},dztag{4}\
        &=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}sum_{k=0}^{n-m}binom{2k}{k}2^{n-m-k},dztag{5}\
        &=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}[t^{n-m}]frac{1}{(1-2t)sqrt{1-4t}},dztag{6}\
        &=int_{0}^1sum_{m=0}^inftybinom{2n-2m}{n-m}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{7}\
        &=int_{0}^inftysum_{m=0}^infty[u^{n-m}]frac{1}{sqrt{1-4u}}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{8}\
        &=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1sum_{m=0}^infty(zu)^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{9}\
        &=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1frac{1}{(1-2zu)sqrt{1-4zu}},dztag{10}\
        &=[u^n]frac{1}{sqrt{1-4u}}left.frac{arctanleft(sqrt{1-4zu}right)}{u}right|_{z=0}^{z=1}tag{11}\
        &,,color{blue}{=[u^{n+1}]frac{1}{sqrt{1-4u}}left(-arctanleft(sqrt{1-4u}right)+frac{pi}{4}right)}tag{12}
        end{align*}

        and we also get when deriving generating functions directly from (1) the same function as OP. The scaling factor $4$ in $sqrt{1-4u}$ is from formula (2) and indicates the connection between $q_n$ and $r_n$ as stated at the beginning of this post.




        Comment:




        • In (4) we use $frac{1}{p+1}=int_0^1z^{p},dz$ where $pne -1$.


        • In (5) we do a rearrangement only.


        • In (6) we apply the coefficient of operator by using (2) and (3) with $p=2$.


        • In (7) we change the order of summation by $mto n-m$ and we replace the upper index $n$ by $infty$ without changing anything, since $binom{2n-2m}{n-m}=0$ when $m>n$.


        • In (8) we apply again the coefficient of operator to $binom{2n-2m}{n-m}$ according to (2).


        • In (9) we use the linearity of the operators and apply the rule $[u^{p-q}]A(u)=[u^p]u^qA(u)$.


        • In (10) we apply the substitution rule of the coefficient of operator with $t=zu$
          begin{align*}
          A(z)=sum_{m=0}^infty a_m z^m=sum_{m=0}^infty z^m [u^m]A(u)
          end{align*}


        • In (11) we integrate obtaining the $arctan$ function.


        • In (12) we finally evaluate the $arctan$ function at lower and upper limit and apply again the rule $[u^n]frac{1}{u}A(u)=[u^{n+1}]A(u)$ as we did in (10).








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 9:36

























        answered Jan 2 at 9:27









        Markus ScheuerMarkus Scheuer

        62.1k459149




        62.1k459149






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057956%2fprove-lim-n-to-infty-sum-m-0n-sum-k-0n-m-frac2n-m-kn-m1%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Ellipse (mathématiques)

            Quarter-circle Tiles

            Mont Emei