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, $


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

and hence,

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

  • $\begingroup$ I believe this is the "more elegant solution" Bill was alluding to. :) $\endgroup$ – J. M. is a poor mathematician Oct 22 '10 at 1:27
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    $\begingroup$ @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. $\endgroup$ – Bill Dubuque Oct 22 '10 at 3:21

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

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    $\begingroup$ how did you find $\rm g_n$? $\endgroup$ – anon Oct 21 '10 at 20:02
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    $\begingroup$ From knowledge of the theory of summation in finite terms, e.g. see the book A = B $\endgroup$ – Bill Dubuque Oct 21 '10 at 20:13
  • $\begingroup$ 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 $N \to \infty$. $\endgroup$ – Hans Lundmark Oct 21 '10 at 20:25
  • $\begingroup$ @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. $\endgroup$ – Bill Dubuque Oct 21 '10 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 $\displaystyle \sum\limits_{n=1}^{\infty}\frac{\Gamma(x+n)}{\Gamma(y+n)}$


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