 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

By <Stirling's approximation>,
$$
(n!)^{1/n}\sim\frac ne(2\pi n)^{1/(2n)},\qquad
|a_n|^{1/n}=\frac{n^{1+1/n^2}}{(n!)^{1/n}}\longrightarrow e.
$$
Here $(2\pi n)^{1/(2n)}\to1$ and $n^{1/n^2}\to1$, as is seen by taking the <natural logarithm>. The <Cauchy-Hadamard theorem> gives $\boxed{R=e^{-1}}$.

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