TITLE: Evaluate $\sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}\frac{1}{(j+1)^2}$
QUESTION [5 upvotes]: I want to evaluate the sum
$ \displaystyle \sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}\frac{1}{(j+1)^2} $
My approach so far has been the possible use of shifted Legendre polynomial. We know
$Q_n(x)=\displaystyle \sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}x^j$
But I am not being able to relate these two. I see that 
$ \displaystyle \int_0^1Q_n(x)dx=\sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}\frac{1}{j+1}$
but how do I get $\dfrac{1}{(j+1)^2}$ ? I need some hint for this.

REPLY [3 votes]: We can also use the Melzak's identity

$$f\left(x+y\right)=x\dbinom{x+n}{n}\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{f\left(y-k\right)}{x+k},\, x,y\in\mathbb{R},\, x\neq-k$$

(for a reference see Z. A. Melzak, V. D. Gokhale, and W. V. Parker, Advanced Problems and Solutions: Solutions $4458$. Amer. Math. Monthly, $60$ $(1)$ $1953$, $53–54$) which holds for all algebraic polynomials $f$ up to degree $n$. So taking $$f\left(z\right)=\frac{\dbinom{2n-z}{n}}{1+n-z},\, y=n$$ and observing that $$\dbinom{n+k}{k}=\dbinom{n+k}{n}
 $$ we have $$\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\dbinom{n+k}{n}\frac{1}{\left(x+k\right)\left(k+1\right)}=\frac{\frac{\dbinom{n-x}{n}}{{\textstyle 1-x}}}{x\dbinom{x+n}{n}}$$ 
so taking the limit as $x\rightarrow1
 $ we have $$\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\dbinom{n+k}{n}\frac{1}{\left(k+1\right)^{2}}=\lim_{x\rightarrow1}\frac{\frac{\dbinom{n-x}{n}}{{\textstyle 1-x}}}{x\dbinom{x+n}{n}}=\frac{1}{n\left(n+1\right)}
 $$ hence $$\sum_{k=0}^{n}\left(-1\right)^{n+k}\dbinom{n}{k}\dbinom{n+k}{n}\frac{1}{\left(k+1\right)^{2}}=\color{red}{\frac{\left(-1\right)^{n}}{n\left(n+1\right)}}$$ as wanted.