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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

MAS
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    x is a given fixed value it cannot be equal to n – user May 20 '18 at 12:24
  • Do not compare signed series. Instead use the absolute value and compare that to $\sum 1/n^2$. You get absolute convergence. Also, it does not make sense to claim converence if $x \in (-n,n)$, where $n$ is the dummy variable of the series. – GEdgar May 20 '18 at 12:32
  • The series converges for any x, refer also to https://math.stackexchange.com/q/2506837/505767 – user May 20 '18 at 15:50

2 Answers2

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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 $\sum_m ^{\infty}\frac 1 {n^{2}}$.

Kavi Rama Murthy
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  • One term of the series can’t make a series not convergent. These points can be excluded and thus the deries is convergent for any x. – user May 20 '18 at 12:53
  • You are wrong, refer also to https://math.stackexchange.com/q/2506837/505767 – user May 20 '18 at 15:49
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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.

user
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    Won't is cause problem if $x \in \mathbb{Z} \setminus {0}$? – Botond May 20 '18 at 12:29
  • The series is not even defined if $x$ is a non-zero integer. – Kavi Rama Murthy May 20 '18 at 12:41
  • One point is not sufficient to make the series not convergent. We can exclude the values of n for which the series is not defined. – user May 20 '18 at 12:55
  • @Botond You are wrong refer also to https://math.stackexchange.com/q/2506837/505767 – user May 20 '18 at 15:49
  • I'm familiar with excluding the first terms, like in the case of $a_n=\sqrt{n-10}$, but I don't remember seeing the exclusion in the "middle" of the interval. – Botond May 20 '18 at 16:54
  • @Botond Convergence of a series is determined by its behavior when $n\to \infty$. Thus some values can’t be significant whereever they are. – user May 20 '18 at 17:11