TITLE: Radius of convergence of the series $\displaystyle\sum\limits_{n=0}^\infty \frac{n!\,z^{2n}}{(1+n^2)^n}$
QUESTION [5 upvotes]: I am doing the following problem and would like to know whether my answer is correct or not:
Find the Radius of convergence for the complex series $\displaystyle\sum\limits_{n=1}^n \frac{z^{2n}n!}{(1+n^2)^n}$. I found it to be $R<1$. I used the relation $n!=cn^ne^{-n}$ and eventually got $|\frac{z^2}{e(1+n^2)}|$<1 and eventually $|z|<1$. Thus, is my work correct? Any inputs are well appreciated!

REPLY [4 votes]: First we should notice that: The radius of convergence of
$$
\sum_{n=0}^\infty a_nz^{2n}
$$
is $r$ if and only if the radius of convergence of
$$
\sum_{n=0}^\infty a_nz^{n}
$$
is equal to $r^2$.
In our powerseries, with $z^2$ replace by $z$, using the ratio test we have
$$
\frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)!}{(1+(n+1)^2)^{n+1}}}{\frac{n!}{(1+n^2)^{n}}}=
\frac{\frac{n+1}{1+(n+1)^2}}{\left(\frac{1+(n+1)^2}{1+n^2}\right)^n}=\frac{\frac{n+1}{1+(n+1)^2}}{\left(1+\frac{2n+1}{1+n^2}\right)^n}\longrightarrow \frac{0}{\mathrm{e}^2}=0.
$$
Therefore, the radius of convergence of $\displaystyle\sum\limits_{n=1}^n \frac{z^{n}n!}{(1+n^2)^n}$ is $r=\infty$ and so in the radius of convergence of $\displaystyle\sum\limits_{n=1}^n \frac{z^{2n}n!}{(1+n^2)^n}$.