Sum of reciprocal of Pochhamer symbols through multiples of a natural L In a question in StackExchange (https://math.stackexchange.com/questions/4236635/sum-of-quotients-of-gamma-functions), I asked if there is a closed expression for the following sum related with a paper I am working in:
$$\sum_{n=1}^N \frac{\Gamma(Ln)}{\Gamma(Ln+r)},$$
where $L,N$ are positive integers greater than 1 and $r$ is a non-integer with $1<r<2$.
The natural conversion of this sum into an integral, as pointed out in my question, doesn't seem to help at all, since I can't compute it.
I was told that the sum is related to the Fox–Wright function. That is useful in the sense that at least I can give a name to my expression, but of course doesn't help to compute the sum, so I tried to write it as
$$\sum_{n=1}^N \frac{1}{(Ln)_{r}},$$
which is exactly what appears in http://specialfunctionswiki.org/index.php/Sum_of_reciprocal_Pochhammer_symbols_of_a_fixed_exponent with $L=1$. Does somebody know about a generalization of this result?
Anyway, it would suffice for me to compute the infinite sum, that is $$\sum_{n=1}^\infty \frac{\Gamma(Ln)}{\Gamma(Ln+r)},$$
which numerically I see that converges, and to know just the asymptotic expansion of a very similar expression,
$$\sum_{n=1}^N \frac{\Gamma(Ln+1)}{\Gamma(Ln+r)}$$
as $N\to\infty$. This last sum doesn't converge, but I know numerically that its limit when we add some other functions of $N$ exists, so having an asymptotic expansion would be enough.
Thank you so much.
 A: The identity you link on the special functions wiki can be rewritten as
$$\sum_{k=1}^n \frac{\Gamma(k)}{\Gamma(k+r)} = \frac{1}{(r-1)\Gamma(r)} - \frac{n\Gamma(n)}{(r-1)\Gamma(n+r)}$$
This clearly has the form of a telescoping sum: i.e. $$\sum_{k=1}^n (T(k) - T(k-1)) = T(n) - T(0)$$
For $L=1$, then, we have $T(n) = \frac{-\Gamma(n+1)}{(r-1)\Gamma(n+r)}$. With the aid of Wolfram Alpha, we get
$L=2$: $$T(n) = -\frac{\Gamma(2n+2)}{\Gamma(2n+r+2)} {}_3F_2\left(\begin{matrix} 1, n+1, n+\tfrac32 \\ n + \tfrac r2 + 1, n + \tfrac r2 + \tfrac32 \end{matrix} \middle\vert 1\right)$$
$L=3$: $$T(n) = - \frac{\Gamma(3n+3)}{\Gamma(3n+r+3)} {}_4F_3\left(\begin{matrix} 1, n + 1, n + \tfrac43, n + \tfrac53 \\ n + \tfrac r3 + 1, n + \tfrac r3 + \tfrac43, n + \tfrac r3 + \tfrac 53 \end{matrix} \middle\vert 1\right)$$
and in general $$T(n) = - \frac{\Gamma(Ln+L)}{\Gamma(Ln+r+L)} {}_{L+1}F_L\left(\begin{matrix} 1, n + 1, n + \tfrac{L+1}L, \ldots, n + \tfrac{2L-1}L \\ n + \tfrac rL + 1, n + \tfrac rL + \tfrac{L+1}L, \ldots, n + \tfrac rL + \tfrac{2L-1}L \end{matrix} \middle\vert 1\right)$$
So the answer is essentially negative: that sum doesn't seem to generalise in a useful way, because for $L > 1$ the hypergeometric form is really just a reformulation of the Fox-Wright form.
