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Find all x for which $\sum_{n=1}^n x^{n^2} n!$ is convergent

So I tried using the ratio test and got $\lim_{n\to \infty} x^{2n+1}(n+1)$ but I don't know how to proceed from that.

I also tried using the root test which gave me $\lim_{n\to \infty} x^{n}(n!)^{1/n}$ which wasn't really helpful either...

Izzat Zaki
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    The ratio test works perfectly: what is the limit of $(n+1)x^{2n+1}$ when $n\to\infty$? The answer depends on $|x|$... – Did Mar 15 '17 at 06:32

1 Answers1

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The root test actually works well here. Note that

$$(n!)^{1/n}\sim \frac ne$$

as $n\to \infty$. Hence, we see that

$$\lim_{n\to \infty}\left|x^n\right|(n!)^{1/n}=\begin{cases}0&,|x|<1\\\\\infty&,|x|\ge 1\end{cases}$$

And the ratio test works well also since

$$\lim_{x\to \infty}|x|^{2n+1}(n+1)=\begin{cases}0&,|x|<1\\\\\infty&,|x|\ge 1\end{cases}$$

Hence, the series converges for $|x|<1$ and diverges otherwise.

Mark Viola
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