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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!

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    $\begingroup$ $x^{n^2}\ne t^2$ $\endgroup$ – Holo Oct 10 '18 at 14:30
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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 $1$.

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