How to determine the sum of $\sum_{k=1}^{\infty} \frac{k^k}{(k!)^2}$

Question

$$\sum_{k=1}^{\infty} \frac{k^k}{(k!)^2}$$

The series converges by the ratio test but how would I find out the exact sum of the series.

Answer 1

The sum seems to converge, since $k^k \approx \frac{k!e^k}{\sqrt{2 \pi k}}$ using Stirling's approximation. Cancelling out $k!$ the summand becomes $ \frac{e^k}{k!\sqrt{2 \pi k}} < \frac{e^k}{k!} $ for $k \geq 1$. since the latter sum clearly converges to $e^{e}-1$, by comparison test the former sum converges too.

EDIT: it does seem to have a numerical solution too:

http://www.wolframalpha.com/input/?i=sum%28k^k%2F%28%28k!%29^2%29%2Ck%3D1..ifinity%29