# Prove convergence and calculate sum (1/k)^k

I was thinking about this series: $$\sum_{k=1}^{\infty}\left( \frac{1}{k}\right)^k$$ I calculated the limit $$\lim_{k \rightarrow \infty} \left(\frac{1}{k}\right)^k$$ and the result was $0$, so i got the necessary condition, but it's not sufficient. I can't think in other ways to prove this converges( altough i'm almost shure it does) and more important than this, calculate this sum. Thanks in advance.

• Hint for convergence: For $k \ge 2$, we have $(1/k)^k \le (1/k)^2$. – Jordan Payette Jan 11 '17 at 18:21
• It can be shown that the value of the sum is equal to $\int_0^1\frac{dx}{x^x}$ by rewriting $x^x=\exp(x\log x)$, expanding the exponential function as a power series and integrating termwise, but I doubt that there is a closed form for the value of the sum. It is approximately equal to 1.29129. – Tom Jan 11 '17 at 18:48

First of all let's consider sum of $\displaystyle \frac{\pi^2}{6} = \sum_{k = 1}^{\infty}\frac{1}{k^2} > \sum_{k = 1}^{\infty}\frac{1}{k^k}$, so by comparasion test this serie is converge.