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.
Best Answer
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 $1\leq\lim_{n\to\infty}n^{1/n(n+1)}\leq n^{1/n}.$