Is there any closed form for $\sum_{k=1}^n \frac{1}{k^k}$?

Question

Is there any closed form for the summation: $$\sum_{k=1}^n \frac{1}{k^k} = ? $$ or at least a tight lower bound?

There are several question about the infinite sum, like this one and other posts linked there. — Martin Sleziak, Jun 21 '16 at 11:50

score 8 · Accepted Answer · answered Aug 28 '13 at 21:16

8

J. Bernoulli showed that

$$\sum_{n=1}^\infty n^{-n}= \int_0^1 x^{-x}\,dx$$

This result is often called the "Sophomore's dream" because it looks too good to be true.

I don't believe there is a form more "closed" than this.

answered Aug 28 '13 at 21:16

Argon

Do you think there is a closed form lower bound? – super delux Aug 28 '13 at 21:21
@majid.khonji I'm not sure if this is what you are looking for but for $n>1$, the sum trivially is $>1$. – Argon Aug 28 '13 at 21:22
You are right. In fact, I am looking for an answer for another question: http://math.stackexchange.com/questions/478516/any-tight-lower-bound-for-this-expression – super delux Aug 28 '13 at 21:26

1 Answers1