While answering this question I came across this curious sum:
$$S_k=\sum_{n=0}^{\infty}\frac{x^n(n+k+1)!}{n!}$$
Which Wolfram Alpha effortlessly evaluates as
$$S_k=\frac{(k+1)!(1-x)^{-k}}{(x-1)^2}$$
The partial sum formula it spat out involved the hypergeometric function ${}_2F_1$, shown below:
$$\sum_{n=0}^{N}\frac{x^n(n+k+1)!}{n!}$$
$$=\frac{(1-x)^{-k-2}}{(N+1)!}\big[(k+1)!(m+1)!-(1-x)^{k+2}x^{N+1}(k+N+2)!{}_2 F_1(1,k+m+3;m+2;x)\big]$$
I did some research on Wolfram Mathworld and Wikipedia about the properties of the hypergeometric function, but didnt get very far. Indeed, the terms in the sequence satisfy the property that the ration between two terms is a rational function of $n$, but the first term isn't $1$ as requested on Mathworld…
Does anyone know how I can connect this sum to a hypergeometric function and thus derive the formula Wolfram Alpha gives?
Best Answer
$$S_k=(k+1)!\sum_{n=0}^{\infty} {n+k+1 \choose k+1}~ x^n$$ $$S_k=(k+1)! \sum_{n=0}^{\infty} {-k-2 \choose n} (-1)^n x^n= (k+1)!~(1-x)^{-k-2},~ |x|<1. $$
Here we have used $${-p \choose m}=(-1)^m {p+m-1 \choose m}$$ and $$(1-x)^{-p}=\sum_{m=0}^{\infty} {-p \choose m} (-1)^m x^m, |x|<1.$$