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.

  • $\begingroup$ 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... $\endgroup$
    – Kelenner
    Sep 9, 2016 at 19:38
  • $\begingroup$ yes this is a possible option. But the resulting integral has to be tackled. $\endgroup$
    – Debashish
    Sep 9, 2016 at 19:42
  • $\begingroup$ 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? $\endgroup$ Sep 9, 2016 at 21:29
  • $\begingroup$ Oh yes .. that was a mistake. I missed the $(-1)^{n+j}$ factor. I am editing it. $\endgroup$
    – Debashish
    Sep 10, 2016 at 13:50

4 Answers 4


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).

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


This matches the result by @JackD'Aurizio.


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*}


  • 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.


  • 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.


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