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

Question

I am interested in the following result \begin{equation*} \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12} \end{equation*} I want to know the fundamentals.

Thankful in advance.

Answer 1

Let $S:=\sum_{k=1}^\infty\,\frac{1}{(2k-1)^2}$. Then, $\frac{\pi^2}{6}=\sum_{k=1}^\infty \,\frac{1}{k^2}=\sum_{r=0}^\infty \,\frac{S}{4^r}=\frac{4}{3}S$. That is, $S=\frac{\pi^2}{8}$. Ergo, $\sum_{n=1}^\infty\,\frac{(-1)^{n-1}}{n^2}=\sum_{k=1}^\infty\,\frac{1}{(2k-1)^2}-2\sum_{k=1}^\infty\,\frac{1}{k^2}=2\left(\frac{\pi^2}{8}\right)-\frac{\pi^2}{6}=\frac{\pi^2}{12}$. Your answer is wrong.

Answer 2

Here are the definitions of the Riemann zeta function and Dirichlet eta function:

$$\zeta(s)=1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+\frac{1}{6^s}\cdots$$

$$\eta(s)=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\frac{1}{5^s}-\frac{1}{6^s}+\cdots$$

The following manipulation relates these two functions:

$$\eta(s)=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\frac{1}{5^s}-\frac{1}{6^s}+\cdots$$

$$=\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\right)-2\left(\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\cdots\right)$$

$$=\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\right)-\frac{2}{2^s}\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\cdots\right) $$

$$=\left(1-\frac{2}{2^s}\right)\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\right)=(1-2^{1-s})\zeta(s) $$

In fact, since $\eta(s)$ has abscissa of convergence ${\rm Re}(s)>0$ whereas $\zeta(s)$ has abscissa of convergence ${\rm Re}(s)>1$, this can be used to provide an analytic continuation of $\zeta(s)$ to the so-called critical strip $0<{\rm Re}(s)<1$. You seek $\eta(2)$, and computing $\zeta(2)=\frac{\pi^2}{6}$ is known famously as the Basel problem. See the link for many solutions to the problem.

Thus $\eta(2)=(1-2^{1-2})\zeta(2)=\frac{\pi^2}{12}$.

Answer 3

One may derive using the residue theorem:

$$\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2+a^2} = \frac12 \left (\frac1{a^2}-\frac{\pi}{a} \operatorname{csch}{\pi a} \right ) $$

Take the limit as $a \to 0$. Use the fact that

$$\operatorname{csch}{z} = \frac1{z} - \frac{z}{6} + O(z^3)$$

and the result that the desired sum is $\pi^2/12$ follows.