Can somebody explain why the sum is $\frac{-\pi^2}{12}$. Can you somehow use the zeta function? If so, how?
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1Euler's sum is $\frac{\pi^2}{6}$. – Stefano Mar 14 '17 at 11:00
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Of course, you are right. – jubueche Mar 14 '17 at 16:06
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Let's add them together:
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}+\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{(-1)^n+1}{n^2}$$
and for odd $n$, $(-1)^n+1=0$ and for even $n$, $(-1)^n+1=2$, so this rewrites to
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}+\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{2}{(2n)^2}$$
which simplifies to
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}+\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac12\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{12}$$
Thus,
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}=\frac{\pi^2}{12}-\frac{\pi^2}{6}=-\frac{\pi^2}{12}$$
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1@JulianBuchel well, the odd terms cancel out, so we're left with only the even terms ($\frac{2}{n^2}$ with $n$ even), that is, $\frac{2}{(2n)^2}$ for $n=1,2,3,\cdots$ – Mar 14 '17 at 18:32
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$$\sum_{i=1}^\infty \dfrac{(-1)^n}{n^2}$$
$$=\sum_{i=1}^\infty \dfrac{1}{(2n)^2}-\sum_{i=1}^\infty \dfrac{1}{(2n-1)^2}$$
$$=\sum_{i=1}^\infty \dfrac{1}{(2n)^2}-\big(\sum_{i=1}^\infty \dfrac{1}{n^2}-\frac14\sum_{i=1}^\infty \dfrac{1}{n^2}\big)$$
$$=\sum_{i=1}^\infty \dfrac{1}{(2n)^2}-\frac34\sum_{i=1}^\infty \dfrac{1}{n^2}$$
$$=\frac14\sum_{i=1}^\infty \dfrac{1}{n^2}-\frac34\sum_{i=1}^\infty \dfrac{1}{n^2}$$
$$=-\frac12\sum_{i=1}^\infty \dfrac{1}{n^2}$$
JMP
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$1/1^2+1/3^2+1/5^2+\dots=1/1^2+1/2^2+1/3^3+\dots-1/4(1/1^2+1/2^2+1/3^2+\dots$ – JMP Mar 15 '17 at 08:51