The above series can be simplified as:
$$\sum_{n=1}^{\infty}\frac{n!}{(k)^n}$$
Does it mean that there is no such a positive integer k exits so that the series converges? If not, what's wrong with my analysis? And how to fix it?
$\endgroup$ 31 Answer
$\begingroup$First we need $a_n \to 0$. We have $a_n = \dfrac{(n!)^2}{(kn)!} \sim \dfrac{2\pi n \left(\dfrac{n}e\right)^{2n}}{\sqrt{2\pi kn} \left(\dfrac{kn}e \right)^{kn}} = \dfrac{\sqrt{2\pi n}}{\sqrt{k}} \dfrac1{k^{kn}} \left(\dfrac{n}{e}\right)^{(2-k)n}$.
Hence, $k \geq 2$ the series converges by comparison test.
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