For which value of $ \ x \in \mathbb{R} \ $ the series $ \ \sum_{n=1}^{\infty} (-1)^n \frac{1}{x^2-n^2} \ $ converges?

Question

For which value of $ \ x \in \mathbb{R} \ $ the series $ \ \sum_{n=1}^{\infty} (-1)^n \frac{1}{x^2-n^2} \ $ converges ?

Answer:

The series is $ \ \sum_{n=1}^{\infty} (-1)^n \frac{1}{x^2-n^2} \ $.

It can be written as $ \ \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2-x^2} \leq \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^2} \ \ \ if \ \ \ x \neq \pm n $

Thus the series converges if $ \ x \in (-n,n) \ $

But I am not sure.

help me

Answer 1

The series converges if and only if $x$ is not an integer or $x=0$. If $x$ is an integer the terms of the series are not all defined, so we have to exclude such $x$. If $x$ is not an integer then there exists $m$ such that $|x| <n^{2}$ for $n \geq m$. Note that $|x^{2}-n^{2}| \geq n^{2}-x^{2}$ so $\sum_m ^{\infty} |\frac {(-1)^n} {x^{2}-n^{2}}| \leq \sum_m ^{\infty}\frac 1 {n^{2}-x^{2}}$. This last series converges by comparison with $\sum_m ^{\infty}\frac 1 {n^{2}}$.

Answer 2

For any $x$ we have that $\frac{1}{n^2-x^2}\to 0$ converges by limit comparison test with $\frac{1}{n^2}$ then the given series converges absolutely and then converges.