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