# convergence radius of power series with $x^{n^2}$

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!

• $x^{n^2}\ne t^2$ – Holo Oct 10 '18 at 14:30

Note tha, 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 $$1$$.