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I am given the following series:

$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$$

I have used the alternating series test to show that the series converges.

However, how do I go about showing what it converges to?

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The homework tag should not be used as the only tag on a question; see tag-wiki and meta. Please include tags pertaining to mathematical content in future questions. –  Lord_Farin Jun 27 '13 at 17:34
    
I think $\sum \frac{1}{n^2}$ is known and $\sum \frac{1}{(2n+1)^2}$ is the Fourier series of some function, or something you can know using Parseval... –  xavierm02 Jun 27 '13 at 17:36
    
Consider the function defined by the series $f(x) = \sum_{n = 1}^\infty \frac{x^n}{n^2}$, and if you it converges at $x = -1$ and you can express it in terms of known functions, then you evaluate $f(-1)$. –  Sammy Black Jun 27 '13 at 17:42
    
@SammyBlack this approach would require the integral of the function $-\frac{\ln(1-x)}{x}$. Maybe this integral is doable, but it looks like it does not have an elementary solution. –  Baby Dragon Jun 27 '13 at 18:18

4 Answers 4

As shown in this answer, $$ \begin{align} \zeta(2) &=\sum_{k=1}^\infty\frac1{k^2}\\ &=\frac{\pi^2}{6} \end{align} $$ Now consider $$ \begin{align} &\hphantom{=\,}\frac1{1^2}\color{#C00000}{-\frac1{2^2}}+ \frac1{3^2}\color{#C00000}{-\frac1{4^2}}+ \frac1{5^2}\color{#C00000}{-\frac1{6^2}}+\dots\\ &=\frac1{1^2}+\frac1{2^2}+ \frac1{3^2}+\frac1{4^2}+ \frac1{5^2}+\frac1{6^2}+\dots\\ &-2\left(\,\,\hphantom{+}\frac1{2^2}\hphantom{+\frac1{3^2}\;}+\frac1{4^2}\hphantom{+\frac1{3^2}\:}+\frac1{6^2}+\dots\right) \end{align} $$ Now note that the stuff in the parentheses is $\frac14$ of the sum above it.

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Is there any link for the proof of that the sum in the parentheses is $1/4$? –  julypraise Jun 27 '13 at 18:45
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@julypraise: not $\frac14$, but $\frac14$ of the sum above it. Notice that each term is $\frac14$ of each term in $\frac1{1^2}+\frac1{2^2}+ \frac1{3^2}+\frac1{4^2}+ \frac1{5^2}+\frac1{6^2}+\dots$ –  robjohn Jun 27 '13 at 18:55

You may be expected to use the fact, not easy to prove, that the sum $\displaystyle\sum_1^\infty \frac{1}{n^2}$ is equal to $\dfrac{\pi^2}{6}$.

The result was first proved by Euler. There are many proofs. In the usual undergraduate curriculum, a student is most likely to meet a proof when first dealing with Fourier series.

Then note that the sum of the squares of the even terms of that series is $\frac{1}{4}$ of the full sum.

That should be enough to evaluate the alternating sum.

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$\sum_{n\geq1}n^{-2}=\frac{\pi^2}{6}$. –  Kirill Jun 27 '13 at 17:39
    
Finger slipped. –  André Nicolas Jun 27 '13 at 17:40
    
You can prove this with the Fourier series for the second Bernoulli polynomial: $\displaystyle{x}^{2}-x+1/6=\frac{1}{\pi^2}\sum _{n=1}^{\infty }{\frac {\cos \left( nx \right) }{{n }^{2}}}$ –  Graham Hesketh Jun 27 '13 at 18:20

Consider the Fourier series of $g(x)=x^2$ for $-\pi<x\le\pi$: $$g(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx)$$

note $b_n=0$ for an even function $g(t)=g(-t)$ and that:

$$a_n=\frac {1}{\pi} \int _{-\pi }^{\pi }\!{x}^{2}\cos \left( nx \right) {dx} =4\,{\frac { \left( -1 \right) ^{n}}{{n}^{2}}},$$

$$\frac{a_0}{2}=\frac {1}{2\pi} \int _{-\pi }^{\pi }\!{x}^{2} {dx} =\frac{1}{3}\pi^2,$$

$$x^2=\frac{1}{3}\,{\pi }^{2}+\sum _{n=1}^{\infty }4\,{\frac { \left( -1 \right) ^{n }\cos \left( nx \right) }{{n}^{2}}},$$ $$x=0\rightarrow \frac{1}{3}\,{\pi }^{2}+\sum _{n=1}^{\infty }4\,{\frac { \left( -1 \right) ^{n }}{{n}^{2}}}=0,$$

$$\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n} }{{n}^{2}}}=-\frac{1}{12}\,{\pi }^{2}.$$

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where did that first statement come from? –  user72195 Jun 27 '13 at 17:54

The series converges absolutely

$ \sum\limits_{n{\rm = 1}}^\infty {\dfrac{1}{{n^2 }}} = \zeta\left( 2 \right) = \dfrac{{\pi ^2 }}{6} $

the given series, to a greater extent, converges to

$ \sum\limits_{n=1}^\infty {\dfrac{{\left( { - 1} \right)^{n + 1} }}{{n^2 }}} = \dfrac{{\pi ^2 }}{{12}} $

Fourier cosine series for $f(t)=\dfrac{\pi^2}{12}-\dfrac{t^2}{4}$ gives

$ \dfrac{{\pi ^2 }}{{12}} - \dfrac{{t^2 }}{4} = \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n + 1} } \dfrac{{\cos \left( {nt} \right)}}{{n^2 }} $

plugging $t=0$ we get

$ \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n + 1} } \dfrac{1}{n^2 } = \dfrac{{\pi ^2 }}{{12}} $

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