# Solving a sum using the hypergeometric function

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?

• If your ultimate quest is to evaluate $S_k$, then we do not need its partial sum. Indeed, using the formula $$\int_{0}^{\infty}t^ne^{-st}\,\mathrm{d}t=\frac{n!}{s^{k+1}}$$ valid for $k >-1$ and $s > 0$, for $|x|<1$ we have $$S_k=\sum_{n=0}^{\infty}\frac{x^n}{n!}\int_{0}^{\infty}t^{n+k+1}e^{-t}\,\mathrm{d}t=\int_{0}^{\infty}t^{k+1}e^{xt}e^{-t}\,\mathrm{d}t=\frac{(k+1)!}{(1-x)^{k+2}}.$$ Alternatively, this formula can be derived recursively from $$S_{-1}=\frac{1}{1-x},\qquad\frac{\mathrm{d}}{\mathrm{d}x}S_k=S_{k+1}.$$ Jul 5, 2020 at 2:05
• Alternatively, but along the same line as SangchulLee's comment, note that $$S_k = (k+1)! \sum_{n=0}^\infty \binom{n+k+1}{n} x^n = (k+1)! \sum_{n=0}^\infty \binom{-k-2}{n}(-x)^n$$ and then use the binomial theorem. Jul 5, 2020 at 2:10
• @Sangchul Lee Fantastic work! However, I'm a bit confused as to how you got rid of the summation and simply put an $e^{xt}$ into the integral on your second to last line. The recursive definition is perfectly clear though. Jul 5, 2020 at 2:21

$$S_k=(k+1)!\sum_{n=0}^{\infty} {n+k+1 \choose k+1}~ x^n$$ $$S_k=(k+1)! \sum_{n=0}^{\infty} {-k-2 \choose n} (-1)^n x^n= (k+1)!~(1-x)^{-k-2},~ |x|<1.$$
Here we have used $${-p \choose m}=(-1)^m {p+m-1 \choose m}$$ and $$(1-x)^{-p}=\sum_{m=0}^{\infty} {-p \choose m} (-1)^m x^m, |x|<1.$$