<s>[INST] Evaluate $\sum_{j=0}^n(-1)^{n+j}{n\choose j}{{n+j}\choose j}\frac{1}{(j+1)^2}$ 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.
 [/INST] Suppose we seek a closed form of the sum
$$\sum_{q=0}^n (-1)^{n+q} {n\choose q} {n+q\choose q} \frac{1}{(q+1)^2}.$$
This is
$$\sum_{q=0}^n (-1)^{n+q} \frac{q+1}{n+1}
 {n+1\choose q+1} {n+q\choose q} \frac{1}{(q+1)^2}
\\ = \frac{1}{n+1} \sum_{q=0}^n (-1)^{n+q} 
 {n+1\choose q+1} {n+q\choose q} \frac{1}{q+1}.$$
Observe that
$$[z^q] \frac{1}{(1-z)^{n+1}} = {n+q\choose q}$$ and hence
$$\frac{1}{n} [z^{q+1}] \frac{1}{(1-z)^{n}} = \frac{1}{q+1} {n+q\choose q}.$$
We introduce 
$$\frac{1}{q+1} {n+q\choose q} = 
\frac{1}{n} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+2}} 
\frac{1}{(1-z)^n} \; dz$$
and obtain for the sum
$$\frac{(-1)^n}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z} 
\frac{1}{(1-z)^n} 
\sum_{q=0}^n (-1)^q {n+1\choose q+1} \frac{1}{z^{q+1}}
\; dz
\\ = - \frac{(-1)^n}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z} 
\frac{1}{(1-z)^n} 
\sum_{q=1}^{n+1} (-1)^q {n+1\choose q} \frac{1}{z^q}
\; dz
\\ = - \frac{(-1)^n}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z} 
\frac{1}{(1-z)^n} 
\left(-1 + \left(1-\frac{1}{z}\right)^{n+1}\right)
\; dz.$$
Now the first piece here yields
$$- \frac{(-1)^n}{n(n+1)} \times -1 \times
[z^0] \frac{1}{(1-z)^n} = \frac{(-1)^n}{n(n+1)}.$$
The second piece is
$$ - \frac{(-1)^n}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z} 
\frac{1}{(1-z)^n} \frac{(z-1)^{n+1}}{z^{n+1}}\; dz
\\= \frac{1}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z} 
\frac{1}{(1-z)^n} \frac{(1-z)^{n+1}}{z^{n+1}}\; dz
\\= \frac{1}{n(n+1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} (1-z) \; dz.$$
This vanishes when $n\ge 1,$ which we have assumed anyway.
It follows that the desired answer is
$$\frac{(-1)^n}{n(n+1)}.$$
This matches the result by @JackD'Aurizio.
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