Can anyone explain to me what identities are used to get the solution:
$$\sum_{t=1}^{n} \frac{\Gamma(t+a)}{\Gamma{(t)}}=\frac{n\Gamma(a+n+1)}{(a+1)\Gamma{(n+1)}}$$
This is only the last step of a much longer problem I was assigned, and since I have never covered identities of Gamma functions I am not sure how to show this is correct!
I am willing to upload the whole question if it is required.
$\endgroup$ 31 Answer
$\begingroup$Looks like the only identity you need is
$$\Gamma(t+1) = t \cdot \Gamma(t)$$
Also, it helps to know that you can work out the value of $\Gamma$ for integers with the formula (which follows by induction from the above and from $\Gamma(1)=1$):
$$\Gamma(n) = (n-1)!$$
Working out your problem: For $n=1$, the left hand side is
$$\frac {\Gamma(1+a)} {\Gamma(1)} =\frac {\Gamma(1+a)} {1}= \Gamma(1+a)$$
and the right hand side is
$$\frac {\Gamma(a+2)} {(a+1)\Gamma(2)}$$
Using the identity I mentioned in the beginning, this reduces to
$$\frac {\Gamma(a+2)} {(a+1)\Gamma(2)} = \frac {(a+1)\Gamma(a+1)} {(a+1)\cdot 1} = \Gamma(1+a)$$
You can work the rest by induction on $n$.
$\endgroup$ 6
