I am working with Gamma functions and I needto compute the following sum: \begin{align} \sum_{k=1}^n k\frac{\Gamma(n-k+\alpha)}{\Gamma(n-k+1)}=\sum_{k=1}^nk(n-k+\alpha)(n-k+\alpha-1)\cdots(n-k+1), \end{align} where $\alpha$ is an integer larger than $1$.
Is it possible in your opinion to compute an explicit expression of the previous sum?
Since all arguments are positive integers, we can use factorials. Our sum becomes $$\sum_{k=1}^n \frac{(n-k+\alpha-1)!}{(n-k)!}k$$ Multiply by $\frac{(\alpha-1)!}{(\alpha-1)!}$ to get that our sum is equivalent to $$(\alpha-1)!\sum_{k=1}^n \frac{(n-k+\alpha-1)!}{(n-k)!(\alpha-1)!}k$$ $$(\alpha-1)!\sum_{k=1}^n \binom{n-k+\alpha-1}{\alpha-1}k$$ We can reverse the bounds of summation to get that our sum is equivalent to $$(\alpha-1)!\sum_{k=0}^{n-1} \binom{k+\alpha-1}{\alpha-1}(n-k)$$ Note that $n-k=\sum_{j=k+1}^n 1$, so we can say our sum is equivalent to $$(\alpha-1)!\sum_{k=0}^{n-1} \left(\binom{k+\alpha-1}{\alpha-1}\sum_{j=k+1}^n 1\right)$$ $$(\alpha-1)!\sum_{k=0}^{n-1}\sum_{j=k+1}^n \binom{k+\alpha-1}{\alpha-1}$$ Now, we switch the order of summation to get $$(\alpha-1)!\sum_{j=1}^n \sum_{k=0}^{j-1} \binom{k+\alpha-1}{\alpha-1}$$ By hockey-stick identity, the inner sum is $$(\alpha-1)!\sum_{j=1}^n \binom{j+\alpha-1}{\alpha}$$ Applying hockey-stick identity again, we get $$(\alpha-1)!\binom{n+\alpha}{\alpha+1}$$
You can also use this identity from here to solve even more complex sums.