Prove $\displaystyle\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(2n+1)^2}=0$

Question

I am trying to prove that the following sum equals to zero: $$\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(2n+1)^2}=0$$ This actually surprised me a-lot. Usually when these kinds of infinite sums equal to zero, it's because of some sort of symmetry, or something like that. But I can't see it here. The $n=0$ term isn't zero, and we don't have $n\mapsto -n$ symmetry. The only thing I thought of is that maybe the Taylor series of $\arctan(x)$ might help, but the sum there starts from $n=1$ (or $n=0$ if you look at $\arctan(x)/x^2$), and needs to be integrated. It was kind of complicated (maybe it's not and I missed something).

P.S. - The original thing I need to compute is the following integral $I$, which I proved that is connected to the mentioned sum through the Residue Theorem:

$$I=\int_{0}^{2\pi}e^{i\theta}\sec{(e^{-i\theta})}\text{d}\theta=\frac 8\pi \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{(2n+1)^2}=0$$

I added this because maybe the integral might be useful somehow (maybe easier to compute). Again I searched for symmetry (the integrand is clearly $2\pi$ symmetric, so I changed the integration interval to be $[-\pi,\pi]$, and then hoped that the integrand would be odd or something. But it's not, if I'm not mistaken).

Thanks

Answer 1

Alternatively, $I=-i\oint_{|z|=1}z^{-2}\sec zdz=0$ because $\sec z$ has no poles in the contour, and the $z^{-2}$ factor doesn't contribute either. (The $-$ sign comes from the contour being clockwise.)

Answer 2

Let us check some terms:

• For $n=-1$ and $n=0$ we get the terms of the series: $$ \frac{(-1)^{-1}}{(-2+1)^2}=-\frac 1{1^2} \text{ and } \frac{(-1)^{0}}{(0+1)^2}=+\frac 1{1^2}\ . $$
• For $n=-2$ and $n=1$ we get the terms of the series: $$ \frac{(-1)^{-2}}{(-4+1)^2}=+\frac 1{3^2} \text{ and } \frac{(-1)^{1}}{(2+1)^2}=-\frac 1{3^2}\ . $$
• For $n=-3$ and $n=2$ we get the terms of the series: $$ \frac{(-1)^{-3}}{(-6+1)^2}=-\frac 1{5^2} \text{ and } \frac{(-1)^{2}}{(4+1)^2}=+\frac 1{5^2}\ . $$
• For $n=-4$ and $n=3$ we get the terms of the series: $$ \frac{(-1)^{-4}}{(-8+1)^2}=+\frac 1{7^2} \text{ and } \frac{(-1)^{3}}{(6+1)^2}=-\frac 1{7^2}\ . $$ The cancelling scheme should be clear now.

Answer 3

You can plot the summands: So evidence suggests there are opposing terms, it's just that they're symmetric about $n=-1/2$ not $n=0$. Symbolically, $$\sum_{n=-\infty}^{-1}\frac{(-1)^n}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}.$$ Now just change the index of the first sum, $$-\sum_{n=-\infty}^0\frac{(-1)^n}{(2n-1)^2}+\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}.$$ Hopefully you can see why this equals zero now.

Answer 4

Note that

$$\sum_{n=-\infty}^\infty{(-1)^n\over(2n+1)^2}=\sum_{m\equiv1\text{ mod }4}{1\over m^2}-\sum_{m\equiv3\text{ mod }4}{1\over m^2}$$

and $m\equiv1$ mod $4$ if and only if $-m\equiv3$ mod $4$, so, since $m^2=(-m)^2$, the two sums on the right hand side are equal, hence cancel each other out.