I'm working on a problem in order statistics. I am hoping to obtain a closed form solution for the following infinite sum:
$$A=\sum_{k=0}^{\infty}\frac{\alpha(\alpha+k\beta)^{k-1}e^{-(\alpha+k\beta)}}{k!}f^{k}$$
where $$\alpha=1$$ and $$\beta=0.5.$$
This is the probability generating function for a so-called Generalized Poisson distribution. I have plugged the equation into Mathematica, but it simply returns back the sum. I would like to work out this specific case first, and then proceed to the more general case.
Please note that this particular case must converge. For example, if we compute this for a large maximum value of $k$ say $100$, you get a sensible numerical answer (which is between $0$ and $1$ as this is a probability if $f$ is between $0$ and $1$).
The answer can be expressed in terms of the Lambert $W$-function.
Let $\color{blue}{L(z)=-W_0(-z)/z}$ for $|z|<1/e$; then, for any complex $\lambda$, we have $$L(z)^\lambda=\sum_{n=0}^\infty\lambda(n+\lambda)^{n-1}\frac{z^n}{n!},$$ so that $\color{blue}{A=e^{-\alpha}L(\beta e^{-\beta}f)^{\alpha/\beta}}$ under obvious conditions.
A possible proof of the above is to deal with $\lambda\in\mathbb{Z}_{>0}$ (and then extend to the general case by the observation that the coefficients of both power series are polynomials in $\lambda$) using Lagrange–Bürmann formula which states that, given analytic functions $\phi$ and $\psi$ with $\phi(0)\neq 0$, if we let $f(w)=w/\phi(w)$ then, using the "coefficient-of" notation, $$[z^n]\psi\big(f^{-1}(z)\big)=\frac1n[w^{n-1}]\big(\psi'(w)\phi(w)^n\big).$$ We apply this to $\phi(w)=e^w$ and $\psi(w)=w^\lambda$. Then $f^{-1}(z)=zL(z)$ and, for $n\geqslant\lambda$, $$[z^n]\big(zL(z)\big)^\lambda=\frac1n[w^{n-1}]\big(\lambda w^{\lambda-1}e^{nw}\big)=\frac{\lambda n^{n-\lambda-1}}{(n-\lambda)!}.$$ This is equivalent to $[z^n]L(z)^\lambda=\lambda(n+\lambda)^{n-1}/n!$, exactly as needed.
Other ideas can be seen in Concrete mathematics [Graham & Knuth & Patashnik] ($2$nd edition, section $5.4$ etc) and The Art of Computer Programming vol. $2$ ($3$rd edition, exercise $4.7.22$).