# A closed form for $\sum_{k=0}^n \frac{ (-1)^k {n \choose k}^2}{k+1}$

Mathematica gives $$\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{k+1}= ~_2F_1[-n,-n;2;-1],$$ where $$~_2F_1$$ that is Gauss hypergeometric function. Here the question is: Can one find a simpler closed form for this summation. Recently, the absolute summation for this has been discussed at MSE:

A binomial summation: $\sum_{k=0}^{n} \frac{{n \choose k}^2}{k+1}$

• We have $$\sum_{k=0}^{n} \frac{(-1)^k}{k+1}\binom{n}{k}^2 = \frac{(-1)^{\lfloor n/2\rfloor}}{n+1}\binom{n}{\lfloor n/2\rfloor}.$$ Commented Mar 1, 2020 at 3:18
• Maybe you can mimic the approaches used here. Commented Mar 1, 2020 at 3:34

Use Binomial identity: $$(1+t)^n=\sum_{k=0}^{n} {n \choose k}t^n~~~(1)$$ Integration of (1) from $$t=0$$ to $$t=x$$,gives $$\frac{(1+x)^{n+1}-1}{n+1}= \sum_{k=0}^n {n \choose k}\frac{x^{k+1}}{k+1}~~~(2)$$ Let $$t=-1/x$$ in (1), then $$(-1)^n x^{-n} (1-x)^n=\sum_{k=0}^{n} (-1)^k {n \choose k} x^{-k}~~~~(3)$$ Multiplying (2) and (3) and collecting the terms of $$x^1$$, we get $$\frac{(-1)^n}{n+1} [(1-x^2)^{n}(1+x)-(1-x)^n]= x^n\sum_{k=0}^{n} \frac{(-1)^k {n \choose k}^2}{k+1} x^1+...+...$$ $$\implies S_n=\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{k+1}=[x^{n+1}] \left((-1)^n \frac{ (1-x^2)^{n}(1+x)-(1-x)^n}{n+1}\right)$$ if $$m=n/2]$$, then $$S_n=(-1)^{m} \frac{{n \choose m}}{n+1}.$$

We seek to evaluate

$$\sum_{k=0}^n \frac{(-1)^k}{k+1} {n\choose k}^2.$$

This is

$$\frac{1}{n+1} \sum_{k=0}^n (-1)^{k} {n\choose k} {n+1\choose k+1} = \frac{1}{n+1} \sum_{k=0}^n (-1)^{k} {n\choose k} {n+1\choose n-k} \\ = [z^n] (1+z)^{n+1} \frac{1}{n+1} \sum_{k=0}^n (-1)^{k} {n\choose k} z^k \\ = [z^n] (1+z)^{n+1} \frac{1}{n+1} (1-z)^n = \frac{1}{n+1} [z^n] (1+z) (1-z^2)^n.$$

Now if $$n=2m$$ we get

$$\frac{1}{n+1} [z^{2m}] (1+z) (1-z^2)^n = \frac{1}{n+1} [z^{2m}] (1-z^2)^n \\ = \frac{1}{n+1} [z^{m}] (1-z)^n = \frac{1}{n+1} (-1)^m {n\choose m}.$$

On the other hand when $$n=2m+1$$ we find

$$\frac{1}{n+1} [z^{2m+1}] (1+z) (1-z^2)^n = \frac{1}{n+1} [z^{2m+1}]z (1-z^2)^n \\ = \frac{1}{n+1} [z^{2m}] (1-z^2)^n = \frac{1}{n+1} [z^{m}] (1-z)^n = \frac{1}{n+1} (-1)^m {n\choose m}.$$

We thus have even or odd the closed form

$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{n+1} (-1)^{\lfloor n/2\rfloor} {n\choose \lfloor n/2\rfloor}.}$$

The second case could have been done by inspection given the first. This result matches the comment by @SangchulLee.

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} \sum_{k = 0}^{n}{\pars{-1}^{k}{n \choose k}^{2} \over k + 1} & = \int_{0}^{1}\sum_{k = 0}^{n}{n \choose k}^{2}\pars{-t}^{k}\,\dd t = \int_{0}^{1}\sum_{k = 0}^{n}{n \choose k}\pars{-t}^{k} \bracks{z^{n - k}}\pars{1 + z}^{n}\,\dd t \\[5mm] & = \bracks{z^{n}}\pars{1 + z}^{n}\int_{0}^{1}\sum_{k = 0}^{n} {n \choose k}\pars{-tz}^{k}\,\dd t \\[5mm] & = \bracks{z^{n}}\pars{1 + z}^{n}\int_{0}^{1}\pars{1 - tz}^{n}\,\dd t = \bracks{z^{n}}\pars{1 + z}^{n}\, {\pars{1 - z}^{n + 1} - 1 \over -\pars{n + 1}z} \\[5mm] & = -\,{1 \over n + 1}\bracks{z^{n + 1}}\pars{1 - z^{2}}^{n}\pars{1 - z} \\[5mm] & = {1 \over n + 1}\braces{\bracks{z^{n}}\pars{1 - z^{2}}^{n} - \bracks{z^{n + 1}}\pars{1 - z^{2}}^{n}} \\[5mm] & = \bbx{\left\{\begin{array}{lcl} \ds{{1 \over n + 1}{n \choose n/2}\pars{-1}^{n/2}} & \mbox{if} & \ds{n}\ \mbox{is}\ even \\[3mm] \ds{{1 \over n + 1}{n \choose \bracks{n + 1}/2}\pars{-1}^{\pars{n - 1}/2}} & \mbox{if} & \ds{n}\ \mbox{is}\ odd \end{array}\right.} \\ & \end{align}

One way to look at this is to take:

\begin{align*} \sum_k \frac{(-1)^k}{k + 1} \binom{n}{k}^2 &= \sum_k \frac{(-1)^k}{k + 1} \binom{n}{k} \binom{n}{n - k} \\ &= [z^n] \left( \sum_k \frac{1}{k + 1} \binom{n}{k} z^k \right) \cdot (-1)^n \left( \sum_k (-1)^k \binom{n}{k} z^k \right) \end{align*}

For the pieces, you know that:

\begin{align*} \sum_k \binom{n}{k} z^k &= (1 + z)^n \\ \sum_k \frac{1}{k + 1} \binom{n}{k} z^k &= \frac{1}{z} \sum_k \frac{1}{k + 1} \binom{n}{k} z^{k + 1} \\ &= \frac{1}{z} \int_0^z (1 + t)^n \, d t \\ &= \frac{1}{z} \frac{(1 + z)^{n + 1} - 1}{n + 1} \end{align*}

Thus:

\begin{align*} \sum_k \frac{(-1)^k}{k + 1} \binom{n}{k}^2 &= (-1)^n [z^n] \frac{(1 + z)^{n + 1} - 1}{z (n + 1)} (1 - z)^n \\ &= \frac{(-1)^n}{n + 1} [z^{n + 1}] ((1 + z)^{n + 1} - 1) (1 - z)^n \\ &= \frac{(-1)^n}{n + 1} [z^{n + 1}] ((1 - z^2)^n (1 + z) \\ &= \frac{(-1)^n}{n + 1} \left( [z^n] (1 - z^2)^n [z^{n - 1}] (1 - z^2)^n \right) \\ &= \frac{(-1)^n}{n + 1} \left( (-1)^{n} \binom{2 n}{n} + (-1)^{n - 1} \binom{2 n}{n - 1} \right) \\ &= \frac{1}{n + 1} \left( \binom{2 n}{n} - \binom{2 n}{n - 1} \right) \\ &= \frac{1}{(n + 1)^2} \binom{2 n}{n} \end{align*}

• This result cannot be correct. The sum varies in sign, but your formula is always positive. In particular, consider $n=2$. Commented Mar 1, 2020 at 3:44
• Sorry, the final answer .doesn't seem to be OK.! Commented Mar 1, 2020 at 4:19