Show the value of the series: $sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i$ [on...
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So i have to show the value of the series
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i$$
For this to exist it has to be convergent, so i checked if it is divergent using the limit test:
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=1$$
Which clearly shows, that it is divergent. Now to the problem: Most/all of my fellow students claim to have a seriesvalue. Is the limit test not conclusive evidence for a series to be divergent?
Edit: I am sorry, I messed up. It should be a sum from n to infinity and not from i to n
real-analysis sequences-and-series limits analysis
put on hold as unclear what you're asking by Did, amWhy, Mostafa Ayaz, user10354138, Math_QED 23 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-2
down vote
favorite
So i have to show the value of the series
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i$$
For this to exist it has to be convergent, so i checked if it is divergent using the limit test:
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=1$$
Which clearly shows, that it is divergent. Now to the problem: Most/all of my fellow students claim to have a seriesvalue. Is the limit test not conclusive evidence for a series to be divergent?
Edit: I am sorry, I messed up. It should be a sum from n to infinity and not from i to n
real-analysis sequences-and-series limits analysis
put on hold as unclear what you're asking by Did, amWhy, Mostafa Ayaz, user10354138, Math_QED 23 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type$(-1)^{frac{n(n + 1)}{2}}$
.
– N. F. Taussig
Nov 22 at 10:35
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
So i have to show the value of the series
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i$$
For this to exist it has to be convergent, so i checked if it is divergent using the limit test:
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=1$$
Which clearly shows, that it is divergent. Now to the problem: Most/all of my fellow students claim to have a seriesvalue. Is the limit test not conclusive evidence for a series to be divergent?
Edit: I am sorry, I messed up. It should be a sum from n to infinity and not from i to n
real-analysis sequences-and-series limits analysis
So i have to show the value of the series
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i$$
For this to exist it has to be convergent, so i checked if it is divergent using the limit test:
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=1$$
Which clearly shows, that it is divergent. Now to the problem: Most/all of my fellow students claim to have a seriesvalue. Is the limit test not conclusive evidence for a series to be divergent?
Edit: I am sorry, I messed up. It should be a sum from n to infinity and not from i to n
real-analysis sequences-and-series limits analysis
real-analysis sequences-and-series limits analysis
edited Nov 22 at 10:38
gimusi
92.5k84495
92.5k84495
asked Nov 22 at 10:23
SirHawrk
12
12
put on hold as unclear what you're asking by Did, amWhy, Mostafa Ayaz, user10354138, Math_QED 23 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Did, amWhy, Mostafa Ayaz, user10354138, Math_QED 23 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type$(-1)^{frac{n(n + 1)}{2}}$
.
– N. F. Taussig
Nov 22 at 10:35
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43
add a comment |
There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type$(-1)^{frac{n(n + 1)}{2}}$
.
– N. F. Taussig
Nov 22 at 10:35
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43
There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type
$(-1)^{frac{n(n + 1)}{2}}$
.– N. F. Taussig
Nov 22 at 10:35
There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type
$(-1)^{frac{n(n + 1)}{2}}$
.– N. F. Taussig
Nov 22 at 10:35
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43
add a comment |
2 Answers
2
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oldest
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up vote
0
down vote
accepted
I calculated the value of the series: $S=sumlimits_{k=0}^infty left(1-(-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Form $S$ in the following way:
$S=sumlimits_{k=0}^infty left(frac 13right)^k-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k=frac {3}{2}-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Separate the sum accordance with even and odd $k$:
If k even then $k=2m$, if odd then $k=2m+1$
Substitute them into S we get:
$S=frac {3}{2}-big(sumlimits_{k=0}^infty (-1)^{m}cdot left(frac 13right)^{2m}+sumlimits_{k=0}^infty (-1)^{3m+1}cdot left(frac 13right)^{2m+1}big)$
Repeat the separation for m values:
If m even then $k=2r$, if odd then $k=2r+1$
we have:
$S=frac {3}{2}-sumlimits_{r=0}^infty big[frac{(-1)^{2r}}{3^{4r}}+frac{(-1)^{2r+1}}{3^{4r+2}}+frac{(-1)^{6r+1}}{3^{4r+1}}frac{(-1)^{6r+4}}{3^{4r+3}}big]=frac {3}{2}-sumlimits_{r=0}^infty big[frac{1}{3^{4r}}big(1-frac{1}{9}-frac{1}{3}+frac{1}{27}big)big]$
Summing the geometrical series and perform the simplifications we get that:
$S=frac{9}{10}$
Using this method you can easily determine the value sum from k=0 to n.
add a comment |
up vote
1
down vote
We are dealing with a sum not with a series, it is a series if we are dealing with its limit as $nto infty$.
To evaluate the sum note that
- $(-1)^{frac{i(i+1)}2}=1,-1,-1,1,1,-1,-1,1,ldots$
then for $n=4k$
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i=sum_{i=0}^{k-1} left(frac 13right)^{1+4k}+sum_{i=0}^{k-1} left(frac 13right)^{2+4k}$$
With reference to the series note that
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=0$$
therefore we can't conclude form here that the series diverges.
Otherwise the series clearly converges since
$$sum_{i=0}^infty left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^ile sum_{i=0}^{infty} left(frac 13right)^{i}$$
and the RHS is a convergent geometric series.
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I calculated the value of the series: $S=sumlimits_{k=0}^infty left(1-(-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Form $S$ in the following way:
$S=sumlimits_{k=0}^infty left(frac 13right)^k-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k=frac {3}{2}-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Separate the sum accordance with even and odd $k$:
If k even then $k=2m$, if odd then $k=2m+1$
Substitute them into S we get:
$S=frac {3}{2}-big(sumlimits_{k=0}^infty (-1)^{m}cdot left(frac 13right)^{2m}+sumlimits_{k=0}^infty (-1)^{3m+1}cdot left(frac 13right)^{2m+1}big)$
Repeat the separation for m values:
If m even then $k=2r$, if odd then $k=2r+1$
we have:
$S=frac {3}{2}-sumlimits_{r=0}^infty big[frac{(-1)^{2r}}{3^{4r}}+frac{(-1)^{2r+1}}{3^{4r+2}}+frac{(-1)^{6r+1}}{3^{4r+1}}frac{(-1)^{6r+4}}{3^{4r+3}}big]=frac {3}{2}-sumlimits_{r=0}^infty big[frac{1}{3^{4r}}big(1-frac{1}{9}-frac{1}{3}+frac{1}{27}big)big]$
Summing the geometrical series and perform the simplifications we get that:
$S=frac{9}{10}$
Using this method you can easily determine the value sum from k=0 to n.
add a comment |
up vote
0
down vote
accepted
I calculated the value of the series: $S=sumlimits_{k=0}^infty left(1-(-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Form $S$ in the following way:
$S=sumlimits_{k=0}^infty left(frac 13right)^k-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k=frac {3}{2}-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Separate the sum accordance with even and odd $k$:
If k even then $k=2m$, if odd then $k=2m+1$
Substitute them into S we get:
$S=frac {3}{2}-big(sumlimits_{k=0}^infty (-1)^{m}cdot left(frac 13right)^{2m}+sumlimits_{k=0}^infty (-1)^{3m+1}cdot left(frac 13right)^{2m+1}big)$
Repeat the separation for m values:
If m even then $k=2r$, if odd then $k=2r+1$
we have:
$S=frac {3}{2}-sumlimits_{r=0}^infty big[frac{(-1)^{2r}}{3^{4r}}+frac{(-1)^{2r+1}}{3^{4r+2}}+frac{(-1)^{6r+1}}{3^{4r+1}}frac{(-1)^{6r+4}}{3^{4r+3}}big]=frac {3}{2}-sumlimits_{r=0}^infty big[frac{1}{3^{4r}}big(1-frac{1}{9}-frac{1}{3}+frac{1}{27}big)big]$
Summing the geometrical series and perform the simplifications we get that:
$S=frac{9}{10}$
Using this method you can easily determine the value sum from k=0 to n.
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I calculated the value of the series: $S=sumlimits_{k=0}^infty left(1-(-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Form $S$ in the following way:
$S=sumlimits_{k=0}^infty left(frac 13right)^k-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k=frac {3}{2}-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Separate the sum accordance with even and odd $k$:
If k even then $k=2m$, if odd then $k=2m+1$
Substitute them into S we get:
$S=frac {3}{2}-big(sumlimits_{k=0}^infty (-1)^{m}cdot left(frac 13right)^{2m}+sumlimits_{k=0}^infty (-1)^{3m+1}cdot left(frac 13right)^{2m+1}big)$
Repeat the separation for m values:
If m even then $k=2r$, if odd then $k=2r+1$
we have:
$S=frac {3}{2}-sumlimits_{r=0}^infty big[frac{(-1)^{2r}}{3^{4r}}+frac{(-1)^{2r+1}}{3^{4r+2}}+frac{(-1)^{6r+1}}{3^{4r+1}}frac{(-1)^{6r+4}}{3^{4r+3}}big]=frac {3}{2}-sumlimits_{r=0}^infty big[frac{1}{3^{4r}}big(1-frac{1}{9}-frac{1}{3}+frac{1}{27}big)big]$
Summing the geometrical series and perform the simplifications we get that:
$S=frac{9}{10}$
Using this method you can easily determine the value sum from k=0 to n.
I calculated the value of the series: $S=sumlimits_{k=0}^infty left(1-(-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Form $S$ in the following way:
$S=sumlimits_{k=0}^infty left(frac 13right)^k-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k=frac {3}{2}-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Separate the sum accordance with even and odd $k$:
If k even then $k=2m$, if odd then $k=2m+1$
Substitute them into S we get:
$S=frac {3}{2}-big(sumlimits_{k=0}^infty (-1)^{m}cdot left(frac 13right)^{2m}+sumlimits_{k=0}^infty (-1)^{3m+1}cdot left(frac 13right)^{2m+1}big)$
Repeat the separation for m values:
If m even then $k=2r$, if odd then $k=2r+1$
we have:
$S=frac {3}{2}-sumlimits_{r=0}^infty big[frac{(-1)^{2r}}{3^{4r}}+frac{(-1)^{2r+1}}{3^{4r+2}}+frac{(-1)^{6r+1}}{3^{4r+1}}frac{(-1)^{6r+4}}{3^{4r+3}}big]=frac {3}{2}-sumlimits_{r=0}^infty big[frac{1}{3^{4r}}big(1-frac{1}{9}-frac{1}{3}+frac{1}{27}big)big]$
Summing the geometrical series and perform the simplifications we get that:
$S=frac{9}{10}$
Using this method you can easily determine the value sum from k=0 to n.
answered Nov 22 at 17:30
JV.Stalker
54639
54639
add a comment |
add a comment |
up vote
1
down vote
We are dealing with a sum not with a series, it is a series if we are dealing with its limit as $nto infty$.
To evaluate the sum note that
- $(-1)^{frac{i(i+1)}2}=1,-1,-1,1,1,-1,-1,1,ldots$
then for $n=4k$
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i=sum_{i=0}^{k-1} left(frac 13right)^{1+4k}+sum_{i=0}^{k-1} left(frac 13right)^{2+4k}$$
With reference to the series note that
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=0$$
therefore we can't conclude form here that the series diverges.
Otherwise the series clearly converges since
$$sum_{i=0}^infty left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^ile sum_{i=0}^{infty} left(frac 13right)^{i}$$
and the RHS is a convergent geometric series.
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
add a comment |
up vote
1
down vote
We are dealing with a sum not with a series, it is a series if we are dealing with its limit as $nto infty$.
To evaluate the sum note that
- $(-1)^{frac{i(i+1)}2}=1,-1,-1,1,1,-1,-1,1,ldots$
then for $n=4k$
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i=sum_{i=0}^{k-1} left(frac 13right)^{1+4k}+sum_{i=0}^{k-1} left(frac 13right)^{2+4k}$$
With reference to the series note that
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=0$$
therefore we can't conclude form here that the series diverges.
Otherwise the series clearly converges since
$$sum_{i=0}^infty left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^ile sum_{i=0}^{infty} left(frac 13right)^{i}$$
and the RHS is a convergent geometric series.
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
add a comment |
up vote
1
down vote
up vote
1
down vote
We are dealing with a sum not with a series, it is a series if we are dealing with its limit as $nto infty$.
To evaluate the sum note that
- $(-1)^{frac{i(i+1)}2}=1,-1,-1,1,1,-1,-1,1,ldots$
then for $n=4k$
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i=sum_{i=0}^{k-1} left(frac 13right)^{1+4k}+sum_{i=0}^{k-1} left(frac 13right)^{2+4k}$$
With reference to the series note that
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=0$$
therefore we can't conclude form here that the series diverges.
Otherwise the series clearly converges since
$$sum_{i=0}^infty left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^ile sum_{i=0}^{infty} left(frac 13right)^{i}$$
and the RHS is a convergent geometric series.
We are dealing with a sum not with a series, it is a series if we are dealing with its limit as $nto infty$.
To evaluate the sum note that
- $(-1)^{frac{i(i+1)}2}=1,-1,-1,1,1,-1,-1,1,ldots$
then for $n=4k$
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i=sum_{i=0}^{k-1} left(frac 13right)^{1+4k}+sum_{i=0}^{k-1} left(frac 13right)^{2+4k}$$
With reference to the series note that
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=0$$
therefore we can't conclude form here that the series diverges.
Otherwise the series clearly converges since
$$sum_{i=0}^infty left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^ile sum_{i=0}^{infty} left(frac 13right)^{i}$$
and the RHS is a convergent geometric series.
edited Nov 22 at 11:39
answered Nov 22 at 10:31
gimusi
92.5k84495
92.5k84495
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
add a comment |
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
add a comment |
There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type
$(-1)^{frac{n(n + 1)}{2}}$
.– N. F. Taussig
Nov 22 at 10:35
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43