# Radius of Convergence of a Series

How would I find the radius of convergence of the following series?

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

The ratio test and root test are inconclusive, so I think I have to use the definition of the radius of convergence

$$\frac{1}{R}=\limsup|a_n|^{\frac{1}{n}}.$$

I have been told that the $n$-th coefficient of this series is not $$\frac{(-1)^n}{n}.$$

I am not sure what exactly I should equate to $|a_n|$. Any help is appreciated.

If $|z|\gt 1$, then $\frac{|z|^{n(n+1)}}n$ does not converge to $0$, and when $|z|\lt 1$, $\sum_n |z|^{n(n+1)}$ is convergent, so the radius of convergence is $1$.
Indeed, the $n$-th coefficient is not $\frac{(-1)^n}n$ as if we write the series into the form $\sum_k b_kx^k$, we have $b_{k(k+1)}=\frac{(-1)^k}k$ and $b_n=0$ if $n$ is not of the form $k(k+1)$ for some integer $k$.
• Notice that $n(n+1)$ is an even integer. Oct 20 '13 at 19:39