Finding the sum of an alternating series

I want to find the sum of $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}$$ I know that this is equal to $\frac{\pi^2}{12}$ thus I was thinking this must just be a taylor series of some trigonometric function but after looking it up, I cannot seem to find one that satisfies this. Any suggestions are greatly appreciated.

From below the user states that $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} = \sum_{n=0}^{\infty}\left(\frac{1}{2n+1} \right)^2 -\sum_{n=1}^{\infty}\left(\frac{1}{2n}\right)^2 = \sum_{n=1}^{\infty}\frac{1}{n^2} - 2\sum_{n=1}^{\infty}\left(\frac{1}{2n}\right)^2 = \ldots = \frac{\pi^2}{12}$$

I want to know the details of the $\ldots$ part.

• Notice that: \begin{align}\sum_{n=1}^\infty\frac1{n^2}-2\sum_{n=1}^\infty\left(\frac1{2n}\right)^2&=\sum_{n=1}^\infty\frac1{n^2}-2\sum_{n=1}^\infty\frac1{4n^2}\\&=\sum_{n=1}^\infty\frac1{n^2}-2\times\frac14\sum_{n=1}^\infty\frac1{n^2}\\&=\sum_{n=1}^\infty\frac1{n^2}-\frac12\sum_{n=1}^\infty\frac1{n^2}\\&=\frac12\sum_{n=1}^\infty\frac1{n^2}\\&=\frac12\frac{\pi^2}6\end{align} Commented Jun 18, 2017 at 16:46

2 Answers

The series can be written as $$\sum _0 ^{\infty}\left(\frac {1}{2n+1}\right)^2-\sum _1 ^{\infty}\left(\frac {1}{2n}\right)^2=\sum _1 ^{\infty} \frac {1}{n^2}-2\sum _1 ^{\infty}\left(\frac {1}{2n}\right)^2$$ By adding and subtracting: $\sum _1 ^{\infty} \left(\frac {1}{2n}\right)^2$

Edited part :-

$$\sum _0 ^{\infty}\left(\frac {1}{2n+1}\right)^2+\sum_1 ^{\infty}\left(\frac {1}{2n}\right)^2-2\sum_1 ^{\infty}\left(\frac{1}{2n}\right)^2$$ Now first two sums can be written as $$\sum _1 ^{\infty} \frac {1}{n^2}$$ So the next two can be written as: $$\frac {1}{2}\sum _1 ^{\infty} \frac {1}{n^2}$$ Now we know all the summations. Thus the answer is: $$\frac {\pi ^2}{6}-\frac {\pi^2}{2.6}=\frac {\pi^2}{12}$$

• Could you give a little more detail on the part where you say "adding and subtracting"... Commented Jun 18, 2017 at 16:09
• I have edited the answer. Commented Jun 18, 2017 at 16:21
• I really like your approach but I don't understand how $$\sum_{n=0}^{\infty}\left(\frac{1}{2n+1} \right)^2 -\sum_{n=1}^{\infty}\left(\frac{1}{2n}\right)^2 = \sum_{n=1}^{\infty}\frac{1}{n^2} - 2\sum_{n=1}^{\infty}\left(\frac{1}{2n}\right)^2$$ The rest of your answer is a bit choppy for me and I can't really follow the rest Commented Jun 18, 2017 at 16:22
• See the terms of original series they are as follows $1-\frac {1}{4}+\frac {1}{9}...=1+\frac {1}{9}+\frac {1}{25}-(\frac {1}{4}+\frac {1}{16}+\frac {1}{36})$ now add all terms with squares of even nos. As we have added we need to subtract . Now we have the series as $1+\frac {1}{4}-\frac {1}{4}-\frac {1}{4}+\frac {1}{9}+\frac {1}{16}-\frac {1}{16}-\frac {1}{16}.....$ now group all positive and negative terms we see its $1+\frac {1}{4}+\frac {1}{9}...-2 (\frac {1}{4}+\frac {1}{16}+\frac {1}{36}...)$ now pull out 4 from all negative terms . We see that the series is simply $0.5\frac{1}{n^2}$ Commented Jun 18, 2017 at 16:29
• Ok, thank you. Although how do you get to the $\frac{\pi^2}{6} - \frac{\pi^2}{2\times 6}$ part? Commented Jun 18, 2017 at 16:33

Hint. One may see this as a special value of the dilogarithm function $$\text{Li}_2(z)=\sum_{n=1}^\infty\frac{z^n}{n^2},\qquad |z|\le1,$$ recalling that $$\text{Li}_2(1)=\frac{\pi^2}6$$ one may observe that $$2\left(\text{Li}_2(z)+\text{Li}_2(-z)\right)=\text{Li}_2(z^2)$$ giving, by putting $z=1$, $$\text{Li}_2(-1)=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}=-\frac{\pi^2}{12}.$$

• Its $n+1$ or maybe my answer is wrong. Commented Jun 18, 2017 at 15:59
• Then why are we getting different answers. Can you see my answer if there is a mistake? Commented Jun 18, 2017 at 16:04
• Observe that from my last line one has $$\sum_{n=1}^\infty\frac{(-1)^n}{n^2}=-\frac{\pi^2}{12} \implies \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}$$ as given by the OP. Commented Jun 18, 2017 at 16:06