Series involving power of the index

Question

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.

Answer 1

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 $m-1$.

Answer 2

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.

Answer 3

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 [1] or the other references in [2].

[1] Borel, Émile (1942). "Sur l'emploi du théorème de Bernoulli pour faciliter le calcul d'une infinité de coefficients. Application au problème de l'attente à un guichet". C. R. Acad. Sci. 214: 452–456.

[2] https://en.wikipedia.org/wiki/Borel_distribution