can someone please help me understand how can i find convergence radius for the following power series:
$$ \sum_{n=1}^\infty \frac{x^{n^2}}{2^{n-1} n^n} $$
I tried positioning $t=x^n$ and i found that it converges only for $x=0$, but i'm not sure it's a correct way to approach the question.
thank you very much!
Note that, if $n\in\mathbb N$,$$\sqrt[n]{\left\lvert\frac{x^{n^2}}{2^{n-1}n^n}\right\rvert}=\frac{\lvert x\rvert^n}{2^{1-\frac1n}n}$$and that therefore$$\lim_{n\in\mathbb N}\sqrt[n]{\left\lvert\frac{x^{n^2}}{2^{n-1}n^n}\right\rvert}=\begin{cases}0&\text{ if }\lvert x\rvert<1\\\infty&\text{ if }\lvert x\rvert>1.\end{cases}$$So, the radius of convergence is $1$.