# Evaluating $\sum_{r=1}^{3n-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{3n}r}$

$$\sum_{r = 1}^{3n-1}\left(-1\right)^{r - 1}\,\,\dfrac{r}{{3n \choose r}},\quad n \in 2k,\ k\in \mathbb{Z^+}$$

#### Answer given (much simpler than expected)

$$\dfrac{3n}{3n+2}$$

I tried adding and subtracting 1 to $$r$$ so could use $$\dfrac{\binom{n}r}{r+1}=\dfrac{\binom{n+1}{r+1}}{n+1}$$, but didn't prove to be useful. I know summation and double summation of binomial coefficients to quite to good extent. If you could help...

• Is $n$ assumed to be even? Apr 17, 2020 at 20:31
• @RobPratt edited accordingly, thanks Apr 17, 2020 at 20:35

Let's start by the following : \begin{aligned}\frac{1}{\binom{3n}{r}}=\left(3n+1\right)\int_{0}^{1}{x^{r}\left(1-x\right)^{3n-r}\,\mathrm{d}x}\end{aligned}

Since $$\left(\forall x\in\left[0,1\right]\right)$$, we have : $$\sum_{r=0}^{3n-1}{r\left(-1\right)^{r-1}x^{r}\left(1-x\right)^{3n-r}}=\left(-1\right)^{n+1}x^{3n+1}\left(1-x\right)+3\left(-1\right)^{n}n x^{3n}\left(1-x\right)+x\left(1-x\right)^{3n+1}$$

It can be proven by differentiating the formula giving the sum of consecutive terms of a geometric sequence.

Thus, \begin{aligned}\scriptsize \sum_{r=0}^{3n-1}{\frac{\left(-1\right)^{r-1}r}{\binom{3n}{r}}}&\scriptsize=\left(3n+1\right)\int_{0}^{1}{\sum_{r=0}^{3n-1}{r\left(-1\right)^{r-1}x^{r}\left(1-x\right)^{3n-r}}\,\mathrm{d}x}\\ &\scriptsize=\left(-1\right)^{n+1}\left(3n+1\right)\int_{0}^{1}{x^{3n+1}\left(1-x\right)\mathrm{d}x}+3\left(-1\right)^{n}n\left(3n+1\right)\int_{0}^{1}{x^{3n}\left(1-x\right)\mathrm{d}x}+\left(3n+1\right)\int_{0}^{1}{x\left(1-x\right)^{3n+1}\,\mathrm{d}x}\\ &\scriptsize=\frac{\left(-1\right)^{n+1}\left(3n+1\right)}{\left(3n+2\right)\left(3n+3\right)}+\frac{3\left(-1\right)^{n}n}{3n+2}+\frac{\left(3n+1\right)}{\left(3n+2\right)\left(3n+3\right)} \end{aligned} And hence, if your $$n$$ is even, we can get rid of the $$\left(-1\right)^{n}$$ and get some cancellations to end with : $$\sum_{r=0}^{3n-1}{\frac{\left(-1\right)^{r-1}r}{\binom{3n}{r}}}=\frac{3n}{3n+2}$$

• How did you write the first equation? Apr 17, 2020 at 21:53
• @Zenix, the first equation is from the beta function $$\int_0^1 x^p(1-x)^q\mathrm{d}x=\frac{1}{(p+q+1)\binom{p+q}{p}}$$ Apr 17, 2020 at 22:21

With $$n=2k$$ we have:

\begin{align} \sum_{r=1}^{6k-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{6k}r} &=\sum_{r=1}^{3k-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{6k}r} +(-1)^{3k-1}\frac{3k}{\binom{6k}{3k}} +\sum_{r=3k}^{6k-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{6k}r}\\ &=(-1)^{3k-1}\frac{3k}{\binom{6k}{3k}} +\sum_{r=1}^{3k-1}\left[\dfrac{(-1)^{r-1}\cdot r}{\binom{6k}r} +\dfrac{(-1)^{r-1}\cdot (6k-r)}{\binom{6k}r}\right]\\ &=(-1)^{3k-1}\frac{3k}{\binom{6k}{3k}}+6k\sum_{r=1}^{3k-1}\dfrac{(-1)^{r-1}}{\binom{6k}r}\tag1\\ &=(-1)^{3k-1}\frac{3k}{\binom{6k}{3k}}+6k\frac{6k+1}{6k+2}\left[\frac1{\binom{6k+1}1}+\dfrac{(-1)^{3k-2}}{\binom{6k+1}{3k}}\right]\\ &=\frac{3k}{3k+1}. \end{align} where we used the fact that the sum in (1) is telescopic due to identity: $$\dfrac1{\binom{m}r}=\frac{m+1}{m+2}\left[\frac1{\binom{m+1}r}+\frac1{\binom{m+1}{r+1}}\right].$$

• suggest some sources for the telescopic sum please. Thank you! Apr 17, 2020 at 21:30
• @SaketGurjar The proof is trivial. Just use $\binom mr=\frac{m!}{r!(m-r)!}$.
– user
Apr 17, 2020 at 21:34

This is true more generally if we replace $$3n$$ by $$n$$.

Wolfram|Alpha returns this closed form for the partial sums:

$$\sum_{r=1}^{k-1}\frac{(-1)^{r-1}r}{\binom nr}=\frac{(n+1)\binom nk-(-1)^k\left(k^2(n+2)-k(n^2+3n+3)+n+1\right)}{(n+2)(n+3)\binom nk}\;.$$

This should be provable with induction over $$k$$. Substituting $$k=n$$ yields

$$\sum_{r=1}^{n-1}\frac{(-1)^{r-1}r}{\binom nr}=\frac{(n+1)+(-1)^n(n^2+2n-1)}{(n+2)(n+3)}\;.$$

For even $$n$$, this simplifies to $$\frac n{n+2}$$.