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.
 A: Use Kelenner's hint about 
$$ \int_{0}^{1}x^j(-\log x)\,dx = \frac{1}{(1+j)^2}\tag{1}$$
and exploit the fact that $(-\log x)$ has a nice representation in terms of shifted Legendre polynomials:
$$ (-\log x) = 1+\sum_{j\geq 1}\frac{(-1)^j(2j+1)}{j(j+1)}Q_j(x)\tag{2} $$
since, for any $n\geq 1$,
$$ \int_{0}^{1}(-\log x)Q_n(x)\,dx = \color{red}{\frac{(-1)^n}{n(n+1)}}\tag{3} $$
can be proved through Rodrigues' formula and integration by parts (the derivative of $(-\log x)$ is simple to deal with).
A: 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.
A: Here is another variation of the theme.

The following is valid
  \begin{align*}
\sum_{j=0}^n(-1)^{n+j}\binom{n}{j}\binom{n+j}{j}\frac{1}{(j+1)^2}=\frac{(-1)^n}{n(n+1)}\qquad\qquad n\geq 1
\end{align*}
We obtain for $n\geq 1$
  \begin{align*}
\sum_{j=0}^n&\binom{n}{j}\binom{n+j}{j}\frac{(-1)^{n+j}}{(j+1)^2}\\
&=\frac{(-1)^n}{n(n+1)}\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\binom{n+j}{j+1}\tag{1}\\
&=\frac{(-1)^{n+1}}{n(n+1)}\sum_{j=0}^n\binom{n+1}{j+1}\binom{-n}{j+1}\tag{2}\\
&=\frac{(-1)^{n+1}}{n(n+1)}\sum_{j=1}^{n+1}\binom{n+1}{j}\binom{-n}{j}\tag{3}\\
&=\frac{(-1)^{n}}{n(n+1)}\left(1-\sum_{j=0}^{n+1}\binom{n+1}{j}\binom{-n}{j}\right)\tag{4}\\
\end{align*}

Comment:


*

*In (1) we use
\begin{align*}
\frac{1}{j+1}\binom{n}{j}=\frac{1}{n+1}\binom{n+1}{j+1}\qquad\text{and}\qquad
\frac{1}{j+1}\binom{n+j}{j}=\frac{1}{n}\binom{n+j}{j+1}
\end{align*}

*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (3) we shift the index $j$ to start from $j=1$.

*In (4) we add the term with $j=0$ and subtract $1$ accordingly.
In order to show the  sum in (4) is equal to zero, we use the coefficient of operator $[z^j]$ to denote the coefficient of $z^j$ in a series. This way we can write e.g.
\begin{align*}
[z^j](1+z)^n=\binom{n}{j}
\end{align*}

We obtain
  \begin{align*}
\sum_{j=0}^{n+1}\binom{n+1}{j}\binom{-n}{j}&=\sum_{j=0}^\infty[u^j](1+u)^{n+1}[z^j](1+z)^{-n}\tag{5}\\
&=[u^0](1+u)^{n+1}\sum_{j=0}^\infty u^{-j}[z^j](1+z)^{-n}\tag{6}\\
&=[u^0](1+u)^{n+1}\left(1+\frac{1}{u}\right)^{-n}\tag{7}\\
&=[u^0]u^n(1+u)\tag{8}\\
&=0\tag{9}
\end{align*}
  and the claim follows.

Comment:


*

*In (5) we apply the coefficient of operator twice and set the upper limit of the sum to $\infty$ without changing anything since we are adding zeros only.

*In (6) we do some rearrangements, use the linearity of the coefficient of operator and use the rule $[u^{p+q}]A(u)=[u^p]u^{-q}A(u)$.

*In (7) we apply the substitution rule of  the coefficient of operator with $z=\frac{1}{u}$
\begin{align*}
A(u)=\sum_{j=0}^\infty a_j u^j=\sum_{j=0}^\infty u^j [z^j]A(z)
\end{align*}

*In (8) we do some simplifications.

*In (9) we select the coefficient of $u^0$.
A: 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.
