# Evaluation of $\sum_{x=1}^{\infty}x^{-x}$ [duplicate]

Possible Duplicate:
“Closed” form for $\sum \frac{1}{n^n}$

Is it possible to evaluate this sum, and if so, how would you do it? This question has been irritating me for a while.

$$\sum_{x=1}^{\infty}x^{-x}$$

It clearly converges, as is proved by the comparison test:

$$\sum_{x=1}^{\infty}x^{-x} \le \sum_{x=1}^{\infty}x^{-2}=\pi^2/3!$$

An approximate value of this sum is

$$\sum_{x=1}^{\infty}x^{-x}\approx1.2912859970...$$

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## marked as duplicate by Pedro Tamaroff♦, Byron Schmuland, Kannappan Sampath, TMM, David MitraApr 13 '12 at 17:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

I don't know of any closed form solution, but we do have Sophomore's dream:

$$\displaystyle \int_0^1 x^{-x} \, \text{d}x = \sum_{n=1}^{\infty}n^{-n}$$

The proof of this is an enjoyable exercise.

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