Determine value the following: $L=\sum_{k=1}^{\infty}\frac{1}{k^k}$

Question

Determine value the following: $$L=\sum_{k=1}^{\infty}\frac{1}{k^k}$$

My try:

Put $$a_n=\sum_{k=1}^{n}\frac{1}{k^k}\to a_{n}-a_{n-1}=\frac{1}{n^n}>0$$

$\to \left\{a_n\right\}$ sequence increasing.

And $$a_{n}=\sum_{k=1}^{n}\frac{1}{k^k}=1+\sum_{k=2}^{n}\frac{1}{k^k}<1+\sum_{k=2}^{n}\frac{1}{k^2}<1+\sum_{k=2}^{n}\frac{1}{(k-1)k}=2-\frac{1}{n}<2$$

$\to \left\{a_n\right\}$ sequence converge.

But come here, I don't know how to determine value

Therefore, please help me, I need a solution.

See http://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function — JJacquelin, Dec 26 '13 at 15:13

score 3 · Answer 1 · edited Dec 26 '13 at 15:04

3

$$\sum_{k=1}^{\infty}\dfrac{1}{k^k}=\int_{0}^{1}\dfrac{1}{x^x}dx$$ (Here has a proof.)

edited Dec 26 '13 at 15:04

Shuchang

9,800

answered Dec 26 '13 at 14:40

math110

93,304

Why is that true? – Igor Rivin Dec 26 '13 at 14:43
True, not true: whatever. What's the proof ?! – DonAntonio Dec 26 '13 at 14:48
Why the downvote? – lcv Dec 26 '13 at 14:54
2

why download it?.... – math110 Dec 26 '13 at 15:06

Determine value the following: $L=\sum_{k=1}^{\infty}\frac{1}{k^k}$

1 Answers1