Radius of convergence of $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}z^{n(n+1)}$

Question

I'm interested in finding the radius of convergence of the power series $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}z^{n(n+1)}$. My first step was to rewrite this series into the standard power series form like this:

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}z^{n(n+1)} = \sum_{n=1}^{\infty}b_nz^n $$ where $b_k = \begin{cases} \frac{(-1)^n}{n} & \text{ if } k =n(n+1) \text{ for some } n \in \mathbb{N} \\ 0 &\text{ else} \end{cases} $

We now would like to appeal to the root test. We look at $\lim_{n \to \infty} \left(\frac{1}{n}\right)^{\frac{1}{n(n+1)}} $. Taking natural logs and using L'Hopitals quickly tells us that this limit is 1.

So is the radius of convergence 1? This just doesn't seem right to me.

Answer 1

Hint: Just find $\limsup_{n\to\infty} (|a_{n(n+1)}|^{1/n(n+1)})$ because term that is zero have no contribution to limit superior of $|a_n|^{1/n}.$ Now $\limsup|{a_{n(n+1)}}^{\frac{1}{n(n+1)}}|=(\frac{1}{n})^{1/n(n+1)}=1$ as $1\leq\lim_{n\to\infty}n^{1/n(n+1)}\leq n^{1/n}.$

Answer 2

Both your approach and your answer are correct. If $\vert z \vert \gt 1$, then the terms of the series don't approach $0$ so it can't possibly converge. If $\vert z \vert \lt 1$, then the series is smaller (in absolute value) than $$\sum_{n=1}^\infty {\frac{z^n} n}.$$