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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!

Infiniticism
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    Note that $\frac{(n+1)^2}{(j+1)^2}\binom{n}{j}^2=\binom{n+1}{j+1}^2$ and that $$\sum_{j=0}^n\binom{n+1}{j+1}^2=\sum_{k=0}^{n+1}\binom{n+1}{k}^2-1=\binom{2n+2}{n+1}-1.$$ It remains to show that $$\binom{2n+2}{n+1}=\frac{2^{2n+2}\Gamma\left(\frac{2n+3}{2}\right)}{\sqrt{\pi}(n+1)\Gamma(n+1)},$$ but this doesn't seem difficult. – Batominovski Feb 25 '20 at 06:34

2 Answers2

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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.

Nathanael Skrepek
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Z Ahmed
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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.

zjs
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