# Series involving power of the index

How to prove the following identity $$\sum_{n=1}^{\infty} \frac{n^{n-1} e^{-n}}{n!} = 1$$ analytically (which can be confirmed with $$Mathematica$$)? The standard trick for geometrical series does not seem to work readily.

There are many ways to prove the formula $$\sum_{n=1}^{\infty} \frac{n^{n-1}}{n!}(xe^{-x})^n = x.\tag{1}$$ As Alexandre Eremenko noted, one approach is Lagrange inversion. Another can be found at https://people.math.harvard.edu/~elkies/Misc/abel.pdf.
Here is a sketch of another method that is not too well known. Expanding $$(xe^{-x})^n$$, we find that the coefficient of $$x^m$$ in $$(1)$$ is $$1/m!$$ times $$\sum_{n=1}^m (-1)^{m-n} n^{m-1}\binom{m}{n}.$$ For $$m>1$$ this sum is $$0$$ since it is the $$m$$th difference of a polynomial of degree $$m-1$$.
Let $$T(z)=\sum_{1}^\infty \frac{n^{n-1}x^n}{n!}.$$ This is known as the exponential generating function of unordered rooted trees, It follows from the Burmann-Lagrange formula (example 1 on p. 3) that this function solves $$T(x)=xe^{T(x)}.$$ Your sum is $$y=T(1/e)$$. So we have to solve the equation $$y=e^{y-1}.$$ This has a root $$y=1$$ which is multiple (of multiplicity $$2$$). This means that the graph of the LHS is tangent to the graph of the RHS at the point $$(1,1)$$. Since the LHS is linear and RHS is convex, our equation has unique solution, namely $$y=1$$, which proves your formula.
• The $\,T(x)\,$ is the exponential generating function of OEIS sequence A000169. May 30 at 1:20
The Borel distribution with parameter $$x \in [0,1]$$ is given by $$p_n= \frac{e^{-x n}(x n)^{n-1}}{n!}\,.$$ for $$n \ge 1$$. This distribution represents the probability that in a Galton Watson branching process with a single progenitor and a Poisson$$(x)$$ offspring distribution, the total size of the population will be $$n$$. The derivation of this distribution, which implies that $$\sum_{n=1}^\infty p_n=1$$, can be found in  or the other references in .