Due to catastrophic cancellation in floating point arithmetic I had to change a formula. Empirically, based on the Tayler series, I was able to develop a new formula that fixes the problem. But I still lack the mathematical derivation. I've already done the majority, but I'm stuck with: $$ \forall_{n,j\in\mathbb{N}}:n=\left(2j+2n+1\right)\sum_{r=1}^{n}\left(j+2r-n\right)\frac{\left(2j+1\right)!}{\left(2j+2r+1\right)!}\frac{\left(2n\right)!}{\left(2n-2r+1\right)!} $$ where j is the iterator in a Taylor series (but not important here).
My question is: How can I proof this equation?
I have already tried reordering the terms and proof by induction. But so far unsuccessful. For hints and ideas, I would be very grateful.
Note that $2(j+2r-n)=(2j+2r+1)-(2n-2r+1)$. So, the RHS of your equality is $$\frac{(2j+1)!(2n)!}{2(2j+2n)!}\sum_{r=1}^{n}\left[\frac{(2j+2n+1)!}{(2j+2r)!(2n-2r+1)!}-\frac{(2j+2n+1)!}{(2j+2r+1)!(2n-2r)!}\right]\\=n\frac{(2j+1)!(2n-1)!}{(2j+2n)!}\sum_{r=1}^{n}\left[\binom{2j+2n+1}{2n-2r+1}-\binom{2j+2n+1}{2n-2r}\right]\\\underset{(r=n-k)}{=}n\binom{2j+2n}{2n-1}^{-1}\sum_{k=0}^{n-1}\left[\binom{2j+2n+1}{2k+1}-\binom{2j+2n+1}{2k}\right]\\=n\binom{2j+2n}{2n-1}^{-1}\sum_{k=0}^{2n-1}(-1)^{k-1}\binom{2j+2n+1}{k}=n,$$ since $\displaystyle\sum_{k=0}^{n}(-1)^k\binom{m}{k}=(-1)^n\binom{m-1}{n}$ (proven, say, by induction on $n$).