# Sum of reciprocals of powers [duplicate]

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

Using Weierstrass theorem (any monotonic and bounded sequence is convergent), we can prove that the sequence $u_{n}=\sum_{k=1}^{n}\frac{1}{k^k}$ with $n$ a positive integer is convergent. But can we actually find its limit ?

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## marked as duplicate by David Speyer, Asaf Karagila, Martin Sleziak, Srivatsan, J. M.Dec 11 '11 at 10:53

There is no known closed form expression. Although there are series related to it which have closed forms. Since $k!\approx e^{-k}k^k\sqrt{2\pi k}$, we can look at things like $\frac{k^k}{k!e^k}$. One such limit is $$\sum_{n=1}^\infty \left(\frac{n^n}{n!e^n}-\frac{1}{\sqrt{2\pi n}}\right)=-\frac{2}{3}-\frac{\zeta\left(\frac{1}{2}\right)}{\sqrt{2\pi}}.$$ – Eric Naslund Nov 27 '11 at 9:50
A related question. If we're hard pressed to find a tidy expression for the infinite sum, I don't see how finding an expression for the finite version is any easier. – J. M. Nov 27 '11 at 10:06
Actually, I think this is a duplicate of 21330. Ismail writes "the sequence $u_n$ ... is convergent. But can we actually find its limit?" This seems to me to be asking abut the infinite sum, which is the subject of question 21330. – David Speyer Nov 27 '11 at 18:16

There is the famous sophmore's dream identity: $$\sum_{k=1}^{\infty} \frac{1}{k^k} = \int_0^1 \frac{dx}{x^x}.$$