Prove the identity $ \sum\limits_{i=k}^{n} \frac{1}{(i-k)! (a+i)_{n-i}}={\frac {a+n}{ \left( a{+} k \right) (n{-}k)! } }.$

Question

Playing with hypergeometric series I have got the identity $$ \sum_{i=k}^{n} \frac{1}{(i-k)! (a+i)_{n-i}}={\frac {a+n}{ \left( a{+} k \right) (n{-}k)! } }, $$ here $(x)_s=x(x+1)\cdots (x+s-1),$ $n, k$ are integers and $a$ is not integer number. Is there any direct way to prove it?

EDIT. My attempt for the case $k=0.$ Let $$S_n=\sum_{i=0}^{n} \frac{1}{i! (a+i)_{n-i}}. $$ Then by direct calculation one may verify that $S_n$ satisfy the recurrence equation $$ (a+1)(n+3) S_{ n+3 }=S_{ n+1} -( n-a ) S_{ n+2 } , $$ with the initial condition $$ S_0 =1,S_1={\frac{a+2}{a+1}},S_2 =\frac 1 2{\frac{a+3}{a+1}}. $$ By solving it we get that $$ S_n= {\frac {a+n}{ a n! } }. $$

I think that it works for any $k$ but I don't like it.

Answer 1

We obtain \begin{align*} \color{blue}{\sum_{i=k}^n}&\color{blue}{\frac{(n-k)!}{(i-k)!(a+i)^{\overline{n+i}}}}\tag{1}\\ &=\sum_{i=k}^n\frac{(n-k)!}{(i-k)!(a+n-1)^{\underline{n-i}}}\tag{2}\\ &=\sum_{i=k}^n\binom{n-k}{i-k}\binom{a+n-1}{n-i}^{-1}\tag{3}\\ &=\sum_{i=0}^{n-k}\binom{n-k}{i}\binom{a+n-1}{n-k-i}^{-1}\tag{4}\\ &=(a+n)\int_0^1\sum_{i=0}^{n-k}\binom{n-k}{i}z^{n-k-i}(1-z)^{a-1+k+i}\,dz\tag{5}\\ &=(a+n)\int_0^1z^{n-k}(1-z)^{a-1+k}\sum_{i=0}^{n-k}\binom{n-k}{i}\left(\frac{1-z}{z}\right)^{i}\,dz\tag{6}\\ &=(a+n)\int_0^1z^{n-k}(1-z)^{a-1+k}\left(1+\frac{1-z}{z}\right)^{n-k}\,dz\tag{7}\\ &=(a+n)\int_0^1(1-z)^{a-1+k}\,dz\\ &\color{blue}{=\frac{a+n}{a+k}} \end{align*} and the claim follows.

Comment:

• In (1) we use the rising factorial $n^{\overline{k}}=n(n+1)\cdots (n+k-1)$.

• In (2) we use the falling factorial $n^{\underline{k}}=n(n-1)\cdots (n-k+1)$.

• In (3) we use the identity $\binom{n}{k}=\frac{n^{\underline{k}}}{k!}$.

• In (4) we shift the index $i$ to start with $i=0$.

• In (5) we write the reciprocal of a binomial coefficient using the beta function \begin{align*} \binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz \end{align*}

• In (6) we do some rearrangements as preparation for the next step.

• In (7) we apply the binomial theorem.

Answer 2

First of all we change the variables $$ S_n\mapsto(n-k)!S_n\\ i\mapsto i-k\\ n-k\mapsto n\\ a+k\mapsto a $$ to bring the expression into a simpler form: $$ S_n:=\sum_{i=0}^n\frac{\binom{n}{n-i}}{\binom{a+n-1}{n-i}}=\frac{a+n}{a}.\tag{1} $$ In what follows we assume $a\not\in\{1-n,2-n,\dots,0\}$.

The equality (1) is obviously valid for $n=0$ and for arbitrary $a$: $S_0=1=\frac{a+0}{a}$. Assuming that it is valid for $n-1$ one obtains that it is valid for $n$ as well: $$ S_n=\sum_{i=0}^n\frac{\binom{n}{n-i}}{\binom{a+n-1}{n-i}}=1+\sum_{i=0}^{n-1}\frac{\binom{n}{n-i}}{\binom{a+n-1}{n-i}}=1+\sum_{i=0}^{n-1}\frac{\frac{n}{n-i}\binom{n-1}{n-i-1}}{\frac{a+n-1}{n-i}\binom{a+n-2}{n-i-1}}\\=1+\frac{n}{a+n-1}\sum_{i=0}^{n-1}\frac{\binom{n-1}{n-1-i}}{\binom{a+n-2}{n-1-i}}= 1+\frac{n}{a+n-1}S_{n-1}\\ \stackrel{I.H}{=}1+\frac{n}{a+n-1}\cdot\frac{a+n-1}{a}=\frac{a+n}{a}. $$

Thus the equality is proved.