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I've got difficulties finding the radius of convergence of $\sum_{n=1}^\infty n!x^{n^2}$. I've already tried presenting it as $\sum_{n=1}^\infty b_n x^n$, where $b_n = (\sqrt{n})!$ if $n$ is a square and $b_n=0$ otherwise. I wanted to use the Cauchy–Hadamard formula and ended up with something like this: $$\frac{1}{R}=\limsup\limits_{n\rightarrow\infty}\sqrt[n]{\lvert b_n \rvert} = \lim\limits_{n\rightarrow\infty}\sqrt[n]{(\sqrt{n})!}$$ And I'm not sure if it's correct. Even if it was, I wouldn't know how to calculate this limit.

Tiamin
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  • For the original series, simply use the ratio test. The ratio test is usually good when factorials are present because there is so much cancellation when dividing two consecutive factorials. – User8128 Nov 29 '20 at 14:15

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The ratio test gives

$$L=\lim_{n\to+\infty}\frac{|a_{n+1}|}{|a_n|}$$

$$=x.\lim_{n\to+\infty}(n+1)e^{2n\ln(|x|)}$$

If $ |x|<1 $, $ L=0 $ it converges.

if $ |x|>1 $ , $L=+\infty$, it diverges.

thus the radius is $ R=1$.

hamam_Abdallah
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