# Summation of $\sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)}$

For $x>0$ and $y>x+1$, how do we prove that $$\sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)} = \frac{x}{y-x-1}$$

This is based on Robert Smith's observation above and Robin Chapman's beta trick in some previous problem.

$\frac{\Gamma{(x+n)}\Gamma{(y-x)}}{\Gamma{(y+n)}} = \int_{0}^{1} t^{x+n-1} (1-t)^{y-x-1} dt$,

summing over $n$ we get,

$\Gamma{(y-x)}\sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \int_{0}^{1} \sum_{n \geq 1} t^{x+n-1} (1-t)^{y-x-1} dt = \int_{0}^{1} t^x (1-t)^{y-x-2}dt,$

or,

$\sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \frac{1}{\Gamma(y-x)}\int_{0}^{1} t^{x+1-1}(1-t)^{y-x-1-1}dt = \frac{\Gamma{(x+1)}\Gamma{(y-x-1)}}{\Gamma{(y-x)}\Gamma(y)} = \frac{\Gamma{(x+1)}}{(y-x-1)\Gamma(y)}$

and hence,

$\frac{\Gamma(y)}{\Gamma(x)} \sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \frac{\Gamma(x+1)}{(y-x-1)\Gamma(x)} = \frac{x}{y-x-1}.$

• I believe this is the "more elegant solution" Bill was alluding to. :) Oct 22, 2010 at 1:27
• @J.M.: Alas, while that's nice, it's not the way that I'm trying to recall. Regarding the above integral representation approach, see my post here for some very powerful techniques. Oct 22, 2010 at 3:21

The sum telescopes since the summand $\rm\ f_n = g_{n+1} - g_n\:$ where $\rm\displaystyle\ g_n = \frac{1-n-y}{y-x-1}\ f_n$

• how did you find $\rm g_n$?
– anon
Oct 21, 2010 at 20:02
• From knowledge of the theory of summation in finite terms, e.g. see the book A = B Oct 21, 2010 at 20:13
• Very pretty, Bill! It might be worth pointing out that the condition $y>x+1$ is needed in order to ensure that $g_N \to 0$ as $N \to \infty$. Oct 21, 2010 at 20:25
• @Chandru1: I suspect that there is a more elegant way to derive the answer. Why don't you wait a day before accepting any answer. Oct 21, 2010 at 20:33

This is not a stand-alone proof like the very nice one provided by Bill, but one can note that the sum is a special case of this identity for the hypergeometric function ${}_2F_1$. Let $a=x$, $b=1$, $c=y$, use $\Gamma(z+1)/\Gamma(z)=z$, and subtract 1 to compensate for the fact that the hypergeometric series starts with $n=0$.

Nice observation Hans. I wanted to derive a very similar sum:

$\displaystyle \sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)}=\sum \limits_{n=1}^{\infty}\displaystyle \frac{x^{(n)}}{y^{(n)}}$ where $x^{(n)}$ is the rising factorial which can be written as $x^{(n)} = \displaystyle \frac{\Gamma(x+n)}{\Gamma(x)}$. Then we have: $\displaystyle \frac{\Gamma(y)}{\Gamma(x)}\sum\limits_{n=1}^{\infty}\frac{\Gamma(x+n)}{\Gamma(y+n)}$.

Apparently, this sum converges to $\displaystyle \frac{\Gamma(x+1)\Gamma(y-x-1)}{\Gamma(y)\Gamma(y-x)}$. After simplication we obtain $\displaystyle \frac{x}{y-x-1}$. However, I haven't been able to derive the result for $\displaystyle \sum\limits_{n=1}^{\infty}\frac{\Gamma(x+n)}{\Gamma(y+n)}$