0
$\begingroup$

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 ?

  • 3
    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 $$\sum_{n=1}^\infty \left(\frac{n^n}{n!e^n}-\frac{1}{\sqrt{2\pi n}}\right)=-\frac{2}{3}-\frac{\zeta\left(\frac{1}{2}\right)}{\sqrt{2\pi}}.$$2011-11-27
  • 0
    [A related question.](http://math.stackexchange.com/questions/21330) 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.2011-11-27
  • 2
    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 21330.2011-11-27

2 Answers 2