# Find the convergence radius for this power series

The Problem: Find the convergence radius of $$\sum_{n=0}^{\infty} \frac{n}{5^{n-1}} z^{\frac{(n)(n+1)}{2}}$$

My attempts to find a solution I apply either the ratio test and end up with this expression:

$$\lim_{n \to \infty} \frac{1}{5} |z|^{(n+1)} =L$$

Since I need $$L<1$$ for the series to converge, my radius of convergence has to be $$|z|<1$$.

Is this correct? Or should I use other method?

Any help will appreciated

Observe we have \begin{align} \sum^\infty_{k=0}a_k z^k \end{align} where \begin{align} a_k = \begin{cases} \frac{n}{5^{n-1}}&\text{ if } k = \frac{n(n+1)}{2},\\ 0 & \text{ otherwise } \end{cases}. \end{align} Then by Cauchy-Hadamard theorem (i.e. root test), we see that \begin{align} \frac{1}{R}=\limsup_{k\rightarrow \infty} \sqrt[k]{|a_k|}. \end{align}
By direct calculation, we see that \begin{align} \sqrt[k]{|a_k|} = \begin{cases} \left(\frac{n}{5^{n-1}}\right)^{\frac{2}{n(n+1)}}&\text{ if } k = \frac{n(n+1)}{2},\\ 0 & \text{ otherwise } \end{cases}. \end{align}
In particular, we see that \begin{align} \lim_{n\rightarrow \infty} \left(\frac{n}{5^{n-1}}\right)^{\frac{2}{n(n+1)}} = 1 \end{align} which means \begin{align} \limsup_{k\rightarrow \infty} \sqrt[k]{|a_k|} = 1. \end{align} Then $$R=1$$.
The ratio test works here and your answer is almost correct, though convergence also occurs when $$|z|=1$$. To see this, note that $$L_{n}\equiv\left|\frac{a_{n+1}}{a_{n}}\right|=\left|\frac{\left(n+1\right)5^{1-(n+1)}z^{(n+1)(n+2)/2}}{n5^{1-n}z^{n(n+1)/2}}\right|=\left(\frac{1}{5n}+\frac{1}{5}\right)\left|z\right|^{n+1}$$ and hence $$\lim_{n}L_{n}=\begin{cases} 0 & \text{if }|z|<1,\\ 1/5 & \text{if }|z|=1,\\ \infty & \text{if }|z|>1. \end{cases}$$