# Infinite sum of $1/n^n$

I'm interested in the sum of what I'll call inverse self powers over integers, namely

$$\sum_{n=1}^{\infty}\frac{1}{n^n}$$

Almost by accident I found that

$$\sum_{n=1}^{\infty}\frac{1}{n^n}=\int_0^1\frac{\mathrm{d}x}{x^x}$$

which is a pretty neat identity, and fairly easy to prove at that. But of course it does not help much given that the integral is no easier to compute than the sum.

Numerically one finds that

$$\sum_{n=1}^{\infty}\frac{1}{n^n}\simeq1.291\,285\,997\ldots$$

Is there any way to link this number to anything else from number theory, special functions, etc.?

• – quid Jan 22 '16 at 21:08
• You mention ease of computation. Note that $n^n$ grows "pretty fast", so you would only really have to compute the first few terms to get an answer accurate up to some desired tolerance. I already get 1.29129 at the 7th partial sum. – parsiad Jan 22 '16 at 21:23
• You're right. I didn't mean numerical computation though. – Vinsanity Jan 22 '16 at 21:23
• – JJacquelin Jan 22 '16 at 21:47