Question: Is there a transformation you can use on a sum$$\sum\limits_{k=0}^{\infty}\text{something}$$To transform it into an expression with the gamma function? In other words, is this possible:$$\sum\limits_{k=0}^{\infty}\text{something}=\dfrac {\Gamma(\text{stuff})\Gamma(\text{stuff})\Gamma(\text{stuff})\cdots}{\Gamma(\text{stuff})\Gamma(\text{stuff})\cdots}$$
While trying to prove an identity, I got to$$\sum\limits_{k=0}^{\infty}\dfrac {\left(\frac 12n+1\right)_k(n)_k(-x)_k(-y)_k}{\left(\frac 12n\right)_k(x+n+1)_k(y+n+1)_k}\dfrac {(-1)^k}{k!}\tag{1}$$ And according to the identity, $(1)$ is equal to$$\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)}{\Gamma(n+1)\Gamma\left(x+y+n+1\right)}\tag{2}$$ So how to I start with $(1)$ and get $(2)$?
For any $k\geq 0$, $\frac{(-1)^k}{k!}$ is exactly the residue of the function $\Gamma(z)$ at $z=-k$.
It follows that the given series can be written as
$$ \sum_{k\geq 0}\frac{(n+2k)\Gamma(n+k)\Gamma(-x+k)\Gamma(-y+k)\Gamma(x+n+1)\Gamma(y+n+1)}{\Gamma(x+n+k+1)\Gamma(y+n+k+1)\Gamma(n+1)\Gamma(-x)\Gamma(-y)}\cdot\frac{(-1)^k}{k!} $$ and we "just" have to show that $$ \sum_{k\geq 0}\frac{(n+2k)\Gamma(n+k)\Gamma(-x+k)\Gamma(-y+k)}{\Gamma(x+n+k+1)\Gamma(y+n+k+1)\Gamma(-x)\Gamma(-y)}\cdot\frac{(-1)^k}{k!}=\Gamma(x+y+n+1) \tag{1}$$ That seems an impossible task through integral transforms or similar techniques, but the Herglotz' trick makes it easy. We may consider the RHS of $(1)$ as a meromorphic function of $x$ for fixed values of $y$ and $n$, locate its poles and compute its residues. We may do the same for every term appearing in the LHS of $(1)$, deriving that both the LHS and the RHS of $(1)$ have the same singularities (simple poles) with the same residues. By proving $(1)$ for a specific value of $x$ like $x=\frac{1}{2}$ we get that $(1)$ holds as an identity as soon as the LHS is a convergent series.
This kind of structure tricks allows us to prove many interesting identities , like Ramanujan's identities for the Eisenstein series $E_4,E_6,E_8$.