Convergence of the series :
$$ \sum_{n=1}^{\infty} n! \frac{k^n}{n^k}$$
I used ratio test for the case $k \ge 0$ and got that for $k \gt 0$ the series is divergent and for $k=0$ the series is convergent. But I can't figure how to approach the $k\lt 0$ case.
I've thought of Leibnitz test for alternating series but couldn't do anything and I can't prove the convergence or even the divergence of this alternating series. Any help would be appreciated.
Best Answer
For $k=-h<0$ we have
$$\left|n! \frac{k^n}{n^k}\right|= h^nn^hn! \to \infty$$
indeed by root test
$$(h^nn^hn! )^\frac1n = h \cdot (\sqrt[n]{n})^h\cdot \sqrt[n]{n!} \to \infty$$
then the given series doesn't converge for $k<0$ since $a_k \not \to 0$.