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

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

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

• no reason to think there is a nice closed-form expression for the sum. Why do you ask? – Will Jagy Feb 5 '13 at 20:47
• Well, its on an old test from one of my classes, and the sum was asked for. I'm correct in saying it converges, at least? – MJBoa Feb 5 '13 at 20:49

## 1 Answer

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

• Not really answering the question, the OP already knows it converges... – L. F. Feb 5 '13 at 21:02
• @L.F.: in the discussion above the OP asks if the sum is convergent – Alex Feb 5 '13 at 21:07