What is the convergence radius of $\sum\limits_{n=1}^{\infty}5^{n^2}x^{n^2}$?

Question

What is the convergence radius of $\sum\limits_{n=1}^{\infty}5^{n^2}x^{n^2}$?

I know I can use either $p=1/{\lim\limits_{k\to\infty}\left\lvert\sqrt{a_k}\right\rvert}$ or $p=\lim\limits_{k\to\infty}\left\lvert \frac{a_k}{a_{k+1}}\right\rvert$ for a power-series$\sum\limits_{k=1}^{\infty} a_kx^k$ but how do I extract the $a_k$ out of this? ($a_k \neq 5^{n^2}$, right?)

Answer 1

If $x\neq0$, then$$\left\lvert\frac{5^{(n+1)^2}x^{(n+1)^2}}{5^{n^2}x^{n^2}}\right\rvert=5^{2n+1}\lvert x\rvert^{2n+1}=\bigl(5\lvert x\rvert\bigr)^{2n+1}.$$Since$$\lim_{n\to\infty}\bigl(5\lvert x\rvert\bigr)^{2n+1}=\begin{cases}\infty&\text{ if }\lvert x\rvert>\frac15\\0&\text{ if }\lvert x\rvert<\frac15,\end{cases}$$the radius of convergence is $\frac15$.

Answer 2

We have $$a_k=\begin{cases}5^{n^2}&\text{if }k=n^2\text{ for some }n\\0&\text{otherwise}\end{cases} $$ Hence we have $a_k=5^k$ for infinitely many $k$.

That makes $$\limsup \sqrt[k]{|a_k|}=5 $$ and so $$p=\frac15.$$