"Closed" form for $\sum \frac{1}{n^n}$ Earlier today, I was talking with my friend about some "cool" infinite series and the value they converge to like the Basel problem, Madhava-Leibniz formula for $\pi/4, \log 2$ and similar alternating series etc.
One series that popped into our discussion was $\sum\limits_{n=1}^{\infty} \frac{1}{n^n}$.
Proving the convergence of this series is trivial but finding the value to which converges has defied me so far. Mathematica says this series converges to $\approx 1.29129$.
I tried Googling about this series and found very little information about this series (which is actually surprising since the series looks cool enough to arise in some context).
We were joking that it should have something to do with $\pi,e,\phi,\gamma$ or at the least it must be a transcendental number :-).
My questions are:


*

*What does this series converge to?

*Does this series arise in any context and are there interesting trivia to be known about this series?


I am actually slightly puzzled that I have not been able to find much about this series on the Internet. (At least my Google search did not yield any interesting results).
 A: Certainly you need to check out Sophomore's Dream.
A: Liouville's theorem implies that if your constant is irrational then it is in fact transcendental. Unfortunately, the trivial way of proving irrationality doesn't work. On second thought, the same problem also prevents us from applying Liouville's theorem!
A: This series turns out to be equal to the integral of $1/x^x$ from $0$ to $1$. Sorry I am bad with the notation needed for math symbols here; but if you rewrite $x^{-x}$ as the exponential of $x\log x$, you can expand it as a taylor series and integrate term by term; to find it equals your sum.
This doesn't give an explicit value; but I think it's a pretty cool identity.
EDIT: Just putting what you said into symbols, probably for the fun of it.
$$\int_0^1 x^{-x} dx=\int_0^1 e^{-x \log x} dx$$
$$ e^{-x \log x} =\sum_{k=0}^\infty (-1)^k \frac{x^k\log^kx}{k!}$$
But since
$$\int_0^1 x^k \log^k x dx =(-1)^k \frac{k!}{(k+1)^{k+1}}$$
we get
$$\int_0^1 x^{-x} dx= \sum_{k=0}^\infty \frac{1}{(k+1)^{k+1}}=\sum_{k=1}^\infty \frac{1}{k^k}$$
