Evaluate $\sum_0^\infty \frac{1}{n^n}$

Question

Courtesy of this xkcd comic I now know that

$$ \sum_{n=1}^\infty \frac{1}{n^n} \approx \ln^e(3) $$

Echoing the views of the comic itself, if I ever find myself taking $\ln^e(x)$ then something has gone horribly horribly wrong. What exactly is it that is going on here then?

I tried attempting a proper solution but it seems beyond me. Here's a wolfram link confirming that this equation is actually valid.

I'm not sure, but maybe this can be approximated by a Riemann sum, followed by evaluating that through integration? As of now I am just tagging it as summation, recreational-mathematics, and approximation. After a solution is posted I will edit in tags to reflect how it was solved.

Answer 1

By the way, $$ \sum_{n=1}^\infty \frac{1}{n^n} \approx 1+\frac{\sin \left(\frac{\pi }{e}\right)}{\pi } $$ seems to be a better approximation.

According to RIES, $$\sum_{n=1}^\infty \frac{1}{n^n} \approx\left(\frac{\phi +\pi }{2}\right)^{\frac{1}{\pi+\frac{1}{4} }}$$ is still better but not as good as the one proposed by JJacquelin.

Next one, please !

Answer 2

Very accurate approximations can be computed thanks to series expansions such as the example given by Claude Leibovici (13 exact digits in case of $S(10)$)

Other methods of numerical calculs leads to a lot of numerical appoximations of various kind. Some examples are compared below.

Many surprising formulas are very easily obtained with the method of experimental calculus described in the paper "Mathématiques expérimentales" pubished on Scribd : http://www.scribd.com/JJacquelin/documents (in French, not translated yet)