I was studying on getting the bound for the expectation of absolute sum of Rademacher RV. Please refer the following link. https://mathworld.wolfram.com/RandomWalk1-Dimensional.html.
I encountered the following expression.
$$\sum_{d=1}^{J}\frac{d}{(J+d)!(J-d)!} = \frac{1}{2J!(J-1)!}$$
for any natural number $J$. It would be greatly appreciated if anybody can help me derive the identity in the above link.
Let $S$ be the sum on the left. We have $$(2J)!S=\sum_{d=1}^J\frac{d(2J)!}{(J+d)!(J-d)!}=\sum_{d=1}^J d\binom{2J}{J-d}=\sum_{i=0}^{J-1}(J-i)\binom{2J}i.$$ where we have substituted $d=J-i$. We now claim that $$\sum_{i=0}^r (J-i)\binom{2J}i=\frac{r+1}2\binom{2J}{r+1}.$$ We'll show this by induction; for the base case of $r=0$, both sides are $J$. For the inductive step, $$\sum_{i=0}^{r+1}(J-i)\binom{2J}i=\sum_{i=0}^r(J-i)\binom{2J}i+(J-r-1)\binom{2J}{r+1}=\frac{2J-r-1}2\binom{2J}{r+1},$$ which equals the desired quantity by comparing ratios of consecutive binomial coefficients. So, applying this at $r=J-1$, $$S=\frac1{(2J)!}\cdot\frac{J}2\binom{2J}J=\frac1{2(J!)(J-1)!},$$ as desired.