In this question, a proof using real analysis is given of the following identity $$ \sum_{n=1}^{\infty} \frac{(n-1)!}{n \prod_{i=1}^{n} (a+i)} = \sum_{k=1}^{\infty} \frac{1}{(a+k)^{2}}$$
Is there a combinatorial proof of this identity? If so, does the proof require that $a$ be a natural number?
The proof below is a WZ method proof.
The pair of functions$G(n,k) = \dfrac{(1)_n}{(k+x+1)_n} \, \dfrac{1}{n^2}, \quad$ $F(n,k) = \dfrac{(1)_n}{(k+x+1)_n} \, \dfrac{1}{(k+x+1)n}$
is a WZ (Wilf and Zeilberger) pair. That is $\, G(n,k+1)-G(n,k) = F(n+1,k)-F(n,k)$.
WZ-pairs are gems because of their interesting properties. For example if $$ \lim_{n \to \infty} F(n,k)=0, \quad \text{and} \quad \lim_{k \to \infty} \sum_{n=1}^{\infty} G(n,k)=0.$$
Wilf and Zeilberger proved that
$$\sum_{n=1}^{\infty} G(n,0) = \sum_{k=1}^{\infty} F(1, k-1).$$
The pair of functions that we are considering satisfies these conditions. Hence
$$ \sum_{n=1}^{\infty} \frac{(1)_n}{n^2(x+1)_n} = \sum_{k=1}^{\infty} \frac{1}{(k+x)^2},$$ which is the identity we wanted to prove.
NOTE: With $(a)_n$ we mean the rising factorial.
