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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

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

1 Answers 1

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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.