Evaluate $\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}$ When testing Mathematica, I accidentally found this equality:
$$\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}=\frac{2^{2 n+2} \Gamma \left(\frac{1}{2} (2 n+3)\right)}{\sqrt{\pi } (n+1)^3 \Gamma (n+1)}-\frac{1}{(n+1)^2}$$
Mathematica gives this closed-form directly, but I've found no proof. I'd like you to help to find the solution. Thanks!
 A: Use Binomial identity:
$$
(1+t)^n=\sum_{j=0}^{n} {n \choose j}t^n.
\tag{1}
$$
Integration of $(1)$ from $t=0$ to $t=x$ gives
$$
\frac{(1+x)^{n+1}-1}{n+1}= \sum_{j=0}^n {n \choose j}\frac{x^{j+1}}{j+1}.\tag{2}
$$
We can change $x$ to $1/x$ in $(2)$ to get
$$
\frac{(1+1/x)^{n+1}-1}{n+1}= \sum_{j=0}^n {n \choose j}\frac{x^{-j-1}}{j+1}
.
\tag{3}
$$
Multiplying $(2)$ and $(3)$ and collecting terms free of $x$ on RHS, we get
$$
\frac{x^{-n-1}\big[(1+x)^{2n+2}-(1+x)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}\big]}{(n+1)^2}=\sum_{j=0}^{n} \frac{{n \choose j}^2}{(j+1)^2} x^0+\dots
$$
Equating the coefficients yields
\begin{align}
S_n = \sum_{j=0}^{n} \frac{{n \choose j}^2}{(j+1)^2}
&= [x^{n+1}]~\frac{[(1+x)^{2n+2}-(1+x)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}]}{(n+1)^2} \\
&= \frac{{2n+2 \choose n+1}-1}{(n+1)^2}.
\end{align}
It may be checked to be nothing but the RHS of OP. 
A: The original expression equals 
$$\sum_{j=0}^n\frac{n!^2}{(j+1)!^2(n-j)!^2}=\sum_{j=0}^n\frac{\binom{n+1}{j+1}^2}{(n+1)^2}=\frac{1}{(n+1)^2}\left(\binom{2n+2}{n+1}-1\right)$$
which should become the expression that Mathematica found after manipulating the Gamma function. 
