For which $ \ x \in \mathbb{R} \ $ the following series converge?

Question

$ (i) \ $ $ \ \large \sum_{n=1}^{\infty} \frac{1}{n} \frac{x^n}{1-x^n} \ $

$ (ii) \ $ $ \sum_{n=1}^{\infty} \large (-1)^{n} \frac{1}{x^2-n^2} \ $

Answer:

(i)

The series is $ \ \large \sum_{n=1}^{\infty} \frac{1}{n} \frac{x^n}{1-x^n} \ $

The series $ \ \large \sum_{n=1}^{\infty} \frac{x^n}{n} \ $ converges on $ \ (-1,1) \ $.

Now $ 1-x^n \geq 1 \ \Rightarrow x^n \leq 0 \Rightarrow x \leq 0 $

Thus on $ \ (-1,0] \ $

$ \ \large \sum_{n=1}^{\infty} \frac{1}{n} \frac{x^n}{1-x^n} \ \leq \sum_{n=1}^{\infty} \frac{x^n}{n} \ $

Thus the series $ \ \large \sum_{n=1}^{\infty} \frac{1}{n} \frac{x^n}{1-x^n} \ $ converges on $ \ (-1,0] \cap (-1,1)=(-1,0] \ $

Am I right ?

If not true , then help me with the appropriate process.

That's better. But is ${x^n\over1-x^n}$ less than $x^n$ for all $n$ if $x\lt0$? — Barry Cipra, May 20 '18 at 01:31

score 1 · Accepted Answer · answered May 20 '18 at 14:45

(i)It is obvious that when $x=-1,1$ the series is not defined.

When $x\neq -1, 1$ let us compute $$\lim_{n\rightarrow\infty}\lvert{\frac{C_n}{C_{n-1}}}\rvert =\lim_{n\rightarrow \infty}|\frac{\frac{1}{n}\frac{x^n}{1-x^n}}{\frac{1}{n-1}\frac{x^{n-1}}{1-x^{n-1}}}|=\lim_{n\rightarrow\infty}\lvert \frac{n-1}{n}\frac{x-x^n}{1-x^n}\rvert =\left\{ \begin{align} 1& &x\in(-\infty,-1)\cup(1,\infty)\\ |x|& &x\in(-1,1) \end{align} \right.$$ Therefore, the series is convergent when $x\in(-1,1)$.

When $x\in(-\infty,-1)\cup(1,\infty)$ the series is approximately $\sum_{n=1}^{\infty}-\frac{1}{n}$ so we should expect that the series is divergent. The detailed analysis is the following. When $x\in(1,\infty)$, $\frac{x^n}{1-x^n}<-1$ so $$\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^n}{1-x^n}<\sum_{n=1}^{\infty}-\frac{1}{n}$$ which is divergent. When $x\in(-\infty,-1)$ it is possible to prove that $C_{2k}+C_{2k+1}<-\frac{1}{2k}-\frac{1}{2k+1}, k\in\mathbb{N}$ and by similar argument the series is also divergent.

(ii)When $x\in\mathbb{Z}$ the series is not defined.

When $x\not\in\mathbb{Z}$, $\lvert(-1)^n\frac{1}{x^2-n^2}\rvert<\frac{2}{n^2}$ when $n$ is sufficiently large and it is known that $\sum_{n=1}^{\infty}\frac{2}{n^2}$ is convergent. Therefore the series is convergent when $x\not\in \mathbb{Z}$.

For which $ \ x \in \mathbb{R} \ $ the following series converge?

1 Answers1