Convergence and sum of a series How could I prove that the following series does converge? 
$$\sum_{k = 1}^{+\infty}\ \frac{e^{k!}}{k^{k!}}$$
And how to determine its total sum? 
I think that that series has to converge, because I took ratio test, $n$-th root test and so on.My problem is to compute the whole sum. Any idea? I tried with Stirling too but seems messy.

The terms of the series equal $(e/k)^{k!}.$ So if $k>6,$ then
$$\frac{e}{k} < \frac{1}{2} \implies \left(\frac{e}{k}\right)^{k!} < \left(\frac{1}{2}\right)^{k!} < \left(\frac{1}{2}\right)^{k}.$$
Since $\sum 1/2^k$ converges, the original series converges by the comparison test.
