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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.

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    $\begingroup$ Are all of the $x$'s in the sum supposed to be $k$'s? $\endgroup$ – Joey Zou Jan 5 '16 at 18:49
  • $\begingroup$ Ops.. Sorry! Yes, I'm going to edit! @JoeyZou $\endgroup$ – Von Neumann Jan 5 '16 at 18:57
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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.

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  • $\begingroup$ We probably had some debates, but don't worry: it's/was my fault. I behave bad sometimes, but I just noticed your answer here, and let me accept it with pleasure :) Thanks! $\endgroup$ – Von Neumann Sep 22 '16 at 22:56
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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.

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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.

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