[Math] For which values of $x$ does this series converge

Question

For which values of $x$ does the series presented below converge?

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

Neither the root test nor the ratio test is of much help – I've tried for awhile now – so any hints would be greatly appreciated.

Thanks.

Answer 1

The ratio test does work here:

$$\begin{align} \left|\frac{a_{n+1}}{a_n}\right| &= \left|\frac{x^{n+1}(1-x^{n+1})}{x^n(1-x^n)}\right|\cdot\frac{n}{n+1} \\ &= \left|x\cdot\frac{1-x^{n+1}}{1-x^n}\right|\cdot\frac{n}{n+1} \end{align}$$

If $|x|<1$ then as $n\to\infty$, $x^n\to 0$ and that ratio tends to $x$.

If $|x|=1$ then the series obviously converges for $x=1$ and diverges for $x=-1$.

If $|x|>1$ then the ratio can be written as

$$\left|x\cdot\frac{\frac 1{x^n+1}-1}{\frac 1{x^n+1}-\frac 1x}\right|\cdot\frac{n}{n+1}$$

and that approaches $x\cdot\frac{0-1}{0-\frac 1x}=x^2$.

In all those cases, the radius of convergence is $1$, including convergence at $x=1$ but not $x=-1$.

(Yes, I know we could have ignored the case $|x|>1$, but I wanted to be consistent here in the approach.)