# 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.

-
My guess is "no." –  Grumpy Parsnip Mar 3 at 21:12
WolframAlpha does evaluate at specific $x$ if you want to play with at least the order of the result -- wolframalpha.com/input/… –  gt6989b Mar 3 at 21:16
You can obtain a bound though. Tight in fact. –  Sabyasachi Mar 3 at 21:16
Do you mean that can $f$ be represented as combination of known fuction? lile $e^x$ ..?? –  mesel Mar 3 at 21:25
@mesel Yes I do. –  tom Mar 3 at 21:36

$\sum_{n=1}^\infty \frac{x^n}{n^n} = x$ 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)

-

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.

-
In case you have any trouble proving that, then check the math appendix on this pdf I would write out the proof, but it's easier to explain with graphs. So. –  Sabyasachi Mar 3 at 21:21
If I understand you correctly, than you are trying to point me to the standard integral test en.wikipedia.org/wiki/Convergence_tests#Integral_test ? Fine but can you evaluate $\int \frac{a^x}{x^x} dx$ ? Plus I know that the series converge. I want to sum it. –  tom Mar 3 at 21:23
@tom I asked one smart friend of mine. Apparently we cannot integrate that. –  Sabyasachi Mar 3 at 21:29
@TMM typo. Correcting it. –  Sabyasachi Mar 3 at 21:30
@DanielLittlewood yes. It was a typo. Updated. –  Sabyasachi Mar 3 at 21:30

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.

-
Can you elaborate more on the last sentence? I don't really understand it. –  tom Mar 3 at 21:39
when we think known function, $sin(x),e^x,ln(x),tan(e^x)...$ they have maclaurin series in the form $a_n/n!$. But your function has form $n!/n^n$ so it just does not fit the picture of known function. Of course,it is not a complete proof just something intuative. –  mesel Mar 3 at 21:56
@tom What mesel is saying is in some sense a combinatorial argument: given two exponential generating functions for the sequences $\{f_n\}$ and $\{g_n\}$ $f(x) = \sum\frac{f_nx^n}{n!}$ and $g(x)=\sum\frac{g_nx^n}{n!}$, then many of the natural manipulations on the coefficients - addition, shifting, multiplication by $n$, convolution, etc - correspond to natural operations on the functions $f()$ and $g()$, and this correspondence goes both ways. So if there were some 'clean' means of getting your series from normal functions... –  Steven Stadnicki Mar 3 at 21:57
...then there would be a relatively clean means of building up the term $\frac{n!}{n^n}$ from these sorts of operations. Since none such is known, it's unlikely that a clean generating function exists. –  Steven Stadnicki Mar 3 at 21:58
Thanks, that sounds reasonable. –  tom Mar 3 at 22:09

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 $\lim_{n\to0}n^n=1$.

-