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.
 A: This is an excellent illustration of the power of the logarithm convergence test: if $x_n > 0 \ \forall n$ and $\lim \frac {\log \frac 1 {x_n}} {\log n} \left\{ \begin{eqnarray} <1 \\ > 1 \end{eqnarray} \right.$ then $\sum x_n \left\{ \begin{eqnarray} \text{converges} \\ \text{diverges} \end{eqnarray} \right.$.
In your problem,
$$\lim \limits \frac {\log \frac {k^{k!}} {{\rm e}^{k!}}} {\log k} = \lim k! \frac {\log k - 1} {\log k} = \infty > 1 ,$$
so according to the above test your series converges.
A: For convergence, use the root test:
$$
\sqrt[k]{  \frac{e^{k!}}{k^{k!}} }=\left( \frac{e}{k} \right) ^{(k-1)!} \to 0
$$
as for $k >7$ you have
$$
0< \left( \frac{e}{k} \right) ^{(k-1)!} < \left( \frac{1}{2} \right) ^{(k-1)!} < \left( \frac{1}{2} \right) ^{k-1}
$$
For the limit, it would surprise me if we can calculate it as a closed form, but maybe I am missing something.
A: 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.
