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 
 A: 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 $R=1$. 
A: 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 
$$
\lim_{n}L_{n}=\begin{cases}
0 & \text{if }|z|<1,\\
1/5 & \text{if }|z|=1,\\
\infty & \text{if }|z|>1.
\end{cases}
$$
