Evaluating the sum $\sum_{k=1}^n \frac{k!}{(k+a)!}$ for a parameter $a$

Question

I am looking for ways to simplify the sum $$\sum_{k=1}^n \frac{k!}{(k+a)!}=\sum_{k=1}^n \frac{\Gamma(k+1)}{\Gamma(k+1+a)}$$ where $n\in\mathbb N$, $a>0$ is a real parameter, and $\Gamma$ denotes the Gamma function.

Answer 1

If $a=1$ you will get the harmonic numbers. Suppose $a\neq1$. Let $t_n=\frac{\Gamma(n+1)}{\Gamma(n+1+a)}$. We are looking for $$s_n=\sum_{k=0}^{n-1} t_k.$$

In view of Gosper's algorithm, I will look for a hypergeometric term $z_n$ satisfying $z_{n+1}-z_n=t_n$. My approach here is described thoroughly in the book A=B (by Marko Petkovsek, Herbert Wilf and Doron Zeilberger.)

Indeed then we have $z_n-z_0=s_n$ and we are done. Let $$y:\mathbb N\to\mathbb R, y(n)=y_n$$ be a rational function such that $z_n = y_n t_n$. Then $y_n$ satisfies $r_n y_{n+1}-y_n=1$, where I have defined $r_n=\frac{t_{n+1}}{t_n}=\frac{n+1}{n+1+a}$.

Indeed this means that we get the recurrence relation $$y_{n+1}=\frac{1+n}{1+n+a}+\frac{1+n}{1+n+a} y_n.$$

This linear recurrence relation of first order can be solved as in here, with a particular solution (note that the initial condition doesn't matter) being $$y_n=\frac{a+n-\frac{(a+1) a \Gamma (a+n+1)}{\Gamma (a+2) \Gamma (n+1)}}{1-a}.$$

Indeed this implies $$z_n = y_n t_n =-\frac{\Gamma (n+1) \left(-\frac{(a+1) a \Gamma (a+n+1)}{\Gamma (a+2) \Gamma (n+1)}+a+n\right)}{(a-1) \Gamma (a+n+1)}=\frac{\frac{a (a+1)}{\Gamma (a+2)}-\frac{(a+n) \Gamma (n+1)}{\Gamma (a+n+1)}}{a-1}$$

and hence $$z_n=\frac{\Gamma (a+n)-\Gamma (a) \Gamma (n+1)}{(a-1) \Gamma (a) \Gamma (a+n)}.$$

In particular, $z_0=0$ and thus $$\bbox[15px,border:1px groove navy]{\sum_{k=0}^{n-1} \frac{k!}{(k+a)!}=s_n=z_n=\frac{\Gamma (a+n)-\Gamma (a) \Gamma (n+1)}{(a-1) \Gamma (a) \Gamma (a+n)}.}$$

We can also deduce $$\bbox[15px,border:1px groove navy]{\sum_{k=1}^n \frac{k!}{(k+a)!}=s_{n+1}-t_0=\frac{\Gamma (a+n+1)-\Gamma (a+1) \Gamma (n+2)}{(a-1) \Gamma (a+1) \Gamma (a+n+1)}.}$$

In particular, by this answer, as $n\to\infty$ for $a>1$ fixed, the sum is $$\frac1{(a-1)\Gamma(a+1)}+O(1/n).$$

Answer 2

Following almost @Maximilian Janisch's steps, I arrived to a different result, namely $$\sum_{k=1}^n \frac{k!}{(k+a)!}=\frac 1{a-1}\left(\frac{1}{\Gamma (a+1)}-\frac{\Gamma (n+2)}{\Gamma (n+a+1)} \right)$$