Recurrence relation between two series The problem asks to prove that $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2(n+1)^2\ldots(n+p-1)^2} = \frac{5p+2}{4(p+1)}\frac{1}{p!^2}-\frac{p(p+1)^3}{4}\cdot\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2(n+1)^2\ldots(n+p+1)^2}$$
where $p$ is a natural number.
How can I show that this equation holds? I think that this follows from a Kummer transformation, which I briefly describe here: usually we want to accelerate the convergence of a given series $a_n$ by means of the following obvious transformation: $$\sum a_n = \gamma C + \sum\left(1-\gamma\frac{c_n}{a_n}\right)a_n$$ where $c_n$ is a convergent series of known sum $C$ such that $a_n/c_n \to \gamma \ne 0 $ (a finite number) as $n\to +\infty$; usually $c_n$ is chosen as much close to $a_n$ as possible, to make the convergence more rapid. In this particular example, I tried $c_n = (n+y)a_n-(n+2+y)a_{n+2}$ (which is a telescoping series, thus $C$ is easy to compute) with $y$ to determine such that the sum on the right hand side of the kummer transformation is close to the right hand side of my equation. This choice for $c_n$ is tyipical in other scenarios, but it doesn't really work here. What should be a proper choice? Also, can I approach the problem from another point? Finally let me point out that this equation is a recurrence relation if we let $T_p$ be the sum of the first series. An interesting problem would be to determine the relation between $T_1$ and $T_{2k+1}$ for example, which leads to other potential transformations...
 A: I know this is totally cheating, but defining
$$S_n=\sum_{i=1}^\infty (-1)^{i-1}\prod_{k=0}^n \frac{1}{(i+k)^2}$$
Mathematica finds
$$S_n=\frac{{}_3 F_2([1,1,1],[2+n,2+n];-1)}{(n+1)!^2}$$
With ${}_pF_q$ being a generalized hypergeometric function:
$${}_pF_q([a_1,...,a_p],[b_1,...,b_q];z)=\sum_{k=0}^\infty \frac{\prod_{i=1}^p \Gamma(a_i+k)\prod_{j=1}^q\Gamma(b_j)}{\prod_{i=1}^p\Gamma(a_i)\prod_{j=1}^q \Gamma(b_j+k)}\frac{z^k}{k!}$$
So perhaps we can use some known properties of this special function to aid us in the proof.
In our case since we have positive integers,
$${}_3F_2([1,1,1],[2+n,2+n];-1)=\sum_{k=0}^\infty \left(\frac{k!(1+n)!}{(n+k+1)!}\right)^2(-1)^k$$
Therefore
$$S_n=\sum_{k=0}^\infty \left(\frac{k!}{(n+k+1)!}\right)^2(-1)^k$$
This is nice, since this sum appears to converge a bit faster than the one we started with (you can experiment with it numerically on Desmos). Mathematica also computes closed forms for all the values of $n$ I've tried:
$$S_0=\frac{\pi^2}{12};S_1=3-4\ln(2);S_2=\frac{21-2\pi^2}{48};S_3=\frac{16\ln(2)-11}{54};S_4=\frac{24\pi^2-235}{27648}$$
I'll keep working on this problem. It's interesting.
More work:
The recurrence we aim to prove is
$$S_{n-1}+\frac{n(n+1)^3}{4}S_{n+1}=\frac{5n+2}{4(n+1)n!^2}$$
Let $C$ be the binomial coefficient:
$$C(a,b)=\frac{a!}{(a-b)!b!}$$
We might notice that
$$\frac{k!}{(n+k+l)!}=\frac{1}{(n+l)!C(n+k+l,n+l)}$$
Thus
$$S_n=\frac{1}{(n+1)!^2}\sum_{k=0}^\infty \frac{(-1)^k}{C(n+k+1,n+1)}$$
Using this definition we can restate the aforementioned recurrence as
$$\sum_{k=0}^\infty \frac{(-1)^k}{C(n+k,n)^2}+\frac{n(n+1)}{4(n+2)^2}\sum_{k=0}^\infty \frac{(-1)^k}{C(n+k+2,n+2)^2}=\frac{5n+2}{4(n+1)}$$
Perhaps now we can use the recursive properties of the binomial coefficient:
$$C(n+k+2,n+2)=C(n+k,n)+2\cdot C(n+k,n+1)+C(n+k,n+2)$$
But this is still quite difficult. I'll keep thinking about it.
