How to calculate radius of convergence of the following series? How can I calculate radius of convergence of the following series?
$$\Large
\sum\limits_{n=0}^\infty \frac{5^{n+1}}{\sqrt[n]{(2n)!}}z^{n}
$$
I tried using D'alembert convergence test but cannot figure out how to calculate.
I know the answer is $\LARGE\frac{1}{5}$

This is what I have got.
$$
1\le((2n)!)^{\frac{1}{n^2}}\le((2n)^{2n})^{\frac{1}{n^2}} = (2n)^{\frac{2}{n}}＝2^{\frac{2}{n}}n^{\frac{2}{n}}\xrightarrow{\scriptscriptstyle n\to\infty}1
$$
Therefore
$$
\sqrt[n]{\frac{5^{n+1}}{\sqrt[n]{(2n)!}}} 
=\frac{5^{\frac{n+1}{n}}}{((2n)!)^{\frac{1}{n^2}}}\xrightarrow{\scriptscriptstyle n\to\infty}\frac{5}{1}
$$
Hence, by the Cauchy-Hadamard formula, the radius of convergence is $\frac{1}{5}$.
