Closed form of $\sum \frac{x^n}{n^n}$ Is there a closed form of this series?
$$
f(x) = \sum_{n=1}^\infty \frac{x^n}{n^n}
$$
I tried few standard tricks how to sum a power series but none of them helped. 
 A: $\sum_{n=1}^\infty \frac{x^n}{n^n} = x $ Sphd$(-x;1)$
But, before saying "that's a joke", read the preamble of the paper : "The Sophomore's Dream Function", http://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function
By the way, this leads to :
$\sum_{n=1}^\infty \frac{x^n}{n^n} = x\int_{0}^1 {t^{-xt}}dt$
(From Eq.6:1 and Eq.1:2)
A: HINT: Given  function $f(x)$ monotonically decreasing on $x \in (m,n)$
$$\sum^n_{x=m}f(x) \le\int^n_m f(x)\,dx \le \sum^{n+1}_{x=m+1} f(x)$$.
Why? Try to approximate the integral using rectangles. When done, check if this integral bounded by summations can be converted to summation bounded by integrals.
A: Since $f$ is differantable,$f(x)=\displaystyle\sum\limits_{n=0}^{\infty} {f^{n}(0)x^n\over n!}$.
Thus,we must have $f(0)=0$ and $f^{n}(0)/ n!=1/n^n$ for $1\leq n \implies f^n(0)={n!/n^n}$
That is why it seem not to have closed form since known function does not have this pattern of derivative at $x=0$.
Since known function has derivative in the form of $a_n/n!$ so it is difucult to obtain $n!/n^n$ by combining them.
A: 
Is there a closed form of this series?

No, there is no closed form for this beautiful expression in terms of elementary functions. However, I've noticed the following hopefully-interesting identity, which I want to share with you: $$f(x)=\sum_{n=0}^\infty\frac{x^n}{n^n}\qquad;\qquad e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\qquad=>\qquad\int_0^\infty\frac{f(x)}{e^x}dx=\sum_{n=0}^\infty\frac{n!}{n^n}$$ where $\lim_{n\to0}n^n=1$.
