# 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.

• You can use that $\int_0^1(-\log u)u^k du=\frac{1}{(k+1)^2}$, this gives you that your sum is $-\int_0^1(\log u)Q_n(u)du$. But I do not know if this can be of some help... Sep 9, 2016 at 19:38
• yes this is a possible option. But the resulting integral has to be tackled. Sep 9, 2016 at 19:42
• I just realized the question title does not match the body of the question. Jack D'Aurizio solved the alternating version, as did I. Should we change the title? Sep 9, 2016 at 21:29
• Oh yes .. that was a mistake. I missed the $(-1)^{n+j}$ factor. I am editing it. Sep 10, 2016 at 13:50

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).

• $Q_n(x)=\dfrac{1}{n!} \dfrac{d^n}{dx^n} (x^2-x)^n$ . Shall I have to use this and integrate by parts ? Kindly elaborate Sep 22, 2016 at 13:52
• I think I can use the orthogonality property of shifted Legendre polynomials Sep 22, 2016 at 13:57

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.

Here is another variation of the theme.

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 (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 $u^0$.

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.