How to prove $\sum_{k=1}^\infty\frac{k^k}{k!}x^k=\frac{1}{2}$ where $x=\frac{1}{3}e^{-1/3}$

Question

How to prove that $$ \sum_{k=1}^\infty\frac{k^k}{k!}x^k=\frac{1}{2}, ~\text{where}~~ x=\frac{1}{3}e^{-1/3}~? $$ I found this sum in my notes, but I don't remember where I got it. Any hints or references would be nice.

Answer 1

Idea, too long for a comment: use the Lambert function. $$W(x) = \sum_{n=1}^\infty\frac{(-n)^{n-1}}{n!}x^n,$$ $$W'(x) = \sum_{n=1}^\infty\frac{(-1)^{n-1}n^n}{n!}x^{n-1} = \sum_{n=1}^\infty\frac{n^n}{n!}(-x)^{n-1}$$ $$\cdots$$

Answer 2

Let's consider $\sum_{k=1}^\infty\frac{k^k}{k!}\big(\frac{1}{ae^{1/a}}\big)^k=\frac{1}{a-1}$ for $a > 1$ (as in the question).

Then $\frac{1}{a-1} = - 1+ \frac{1}{1-1/a} = \sum_{k=1}^\infty (\frac1a)^k$ by the geometric series. So one has to show

$\sum_{k=1}^\infty\Big[-1 + \frac{k^k}{k!}{e^{-k/a}}\Big](\frac{1}{a}\big)^k=0$

Writing some terms

$0 = \Big[-1 + {e^{-1/a}}\Big](\frac{1}{a}\big) + \Big[-1 + \frac{2^2}{2!}{e^{-2/a}}\Big](\frac{1}{a}\big)^2 + \Big[-1 + \frac{3^3}{3!}{e^{-3/a}}\Big](\frac{1}{a}\big)^3 + \cdots$

Expanding the exponentials and writing the terms up to $(1/a)^3$, say, requires to take the terms

$0 = \Big[-\frac{1}{a} + \frac12 (\frac{1}{a})^2 + \cdots \Big](\frac{1}{a}\big) + \Big[-1 + \frac{2^2}{2!}(1-\frac{2}{a} + \cdots)\Big](\frac{1}{a}\big)^2 + \Big[-1 + \frac{3^3}{3!}(1 + \cdots)\Big](\frac{1}{a}\big)^3 + \cdots$

Sorting for powers of $1/a$ gives

$0 = \Big[0 \Big](\frac{1}{a}\big) + \Big[-1 -1 + \frac{2^2}{2!}\Big](\frac{1}{a}\big)^2 + \Big[\frac12 - 2 \frac{2^2}{2!} -1 + \frac{3^3}{3!}\Big](\frac{1}{a}\big)^3 + \cdots$

and all the prefactors are zero, as required. The full solution then comes upon expanding all exponentials. Clearly this requires some more formal work.