Radius of convergence of $\sum_{n=1}^{\infty}\frac{x^n}{1-x^n}$?

Question

I'm struggling to find the radius of converge of the following sum: \begin{equation} \sum_{n=1}^{\infty}\frac{x^n}{1-x^n} \end{equation} When I apply the Dalambre's criterion I get: \begin{equation} \lim_{n\rightarrow\infty}\frac{\frac{x^{n+1}}{1-x^{n+1}}}{\frac{x^n}{1-x^n}}=1 \end{equation} When I apply the other criterion: \begin{equation} \lim_{n\rightarrow\infty}n\left(\frac{\frac{x^{n}}{1-x^{n}}}{\frac{x^{n+1}}{1-x^{n+1}}}-1\right)=0 \end{equation} And i have absolutely no idea how to proceed from here...

Answer 1

Thats not a power series so here we don't talk about radius of convergence. $$R=\lim_{n\rightarrow\infty}\frac{\frac{x^{n+1}}{1-x^{n+1}}}{\frac{x^n}{1-x^n}}=\lim_{n\rightarrow\infty}x\frac{1-x^n}{1-x^{n+1}}$$ If $|x|<1$ $$R=x$$ as $x^n$ goes to zero.

If $|x|>1$ $$R=1$$ The criterion does not decide but the general term of the serie does nor converge to zero, so it cannot be convergent.

For $|x|=1$ is not defined. So the values for $x$ such the series is convergent are $(-1,1) $

Answer 2

Let me start by pointing out that the term "radius of convergence" is usually reserved for power series, and this is not a power series.

Now we investigate for which $x\in\mathbb{R}$ the series converges. We start by applying the divergence test. Now clearly $x\notin\{-1,1\}$ right of the bat, as we cannot have division by zero, and then for $x\in\mathbb{R}\setminus\{-1,1\}$,

$$\lim_{n\to\infty}\frac{x^n}{1-x^n}=\lim_{n\to\infty}\left(\frac{1}{1-x^n}-1\right)=\begin{cases} -1, &\lvert x \rvert >1, \\ 0, &\lvert x \rvert <1, \end{cases}$$

and so we know now immediately that if we have convergence, then we must have that $x\in(-1,1)$. We now investigate these $x$. Indeed we have that, if $\lvert x\rvert<1$,

$$\lim_{n\to\infty}\left\lvert \frac{\frac{x^{n+1}}{1-x^{n+1}}}{\frac{x^{n}}{1-x^{n}}} \right\rvert=\lim_{n\to\infty}\left\lvert \frac{x(1-x^{n})}{1-x^{n+1}} \right\rvert=\lvert x \rvert <1.$$

It follows by the ratio test that the series converges for $x\in(-1,1)$. We can thus conclude that

• if $x\in\{-1,1\}$, the series is undefined,
• if $x\in(-1,1)$, the series converges,
• if $x\in\mathbb{R}\setminus[-1,1]$, the series diverges.

We could thus say (although I disapprove of the terminology as we do not have a power series), that the radius of convergence is $1$.