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Find $$S=\sum_{r=1}^{3n-1}{ (-1)^{r-1}r\over{3n \choose r}},~ \text{if $n$ is even}$$

The answer given to me is ${3n\over3n+2}$ , the main problem I am facing is that the binomial coefficients are in the denominator, and so I can not use any of the usual techniques I used to use , like using integration or different on any binomial series. Although I tried by rewriting the sum in reverse order and then adding it to the original expression, this gave me :

$$2S=3n\sum_{r=1}^{3n-1}{ (-1)^{r-1}\over{3n \choose r}}$$

This is sure simpler than the original problem but still no good , I could not figure out a way to solve the rest .

Could someone please help me in solving thi problem ?

Thanks !

Z Ahmed
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ARROW
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  • Start by rewriting the reciprocal of binomial coefficients as a Beta function $$\frac1{\binom{n}{k}}=(n+1)B(k+1,n-k+1)=(n+1)\int_0^1t^k(1-t)^{n-k},\mathrm{d}t.$$ – user10354138 Jul 25 '20 at 15:05
  • It also helps to use$$\sum_{r=1}^{3n-1}rx^r=\frac{x-3nx^{3n}+(3n-1)x^{3n+1}}{(1-x)^2}$$(which you can prove by induction on $n\ge1$), where $x$ is a function of @user10354138's $t$ you'll guess as you work through it. – J.G. Jul 25 '20 at 15:26
  • You may see https://math.stackexchange.com/questions/3251604/proving-that-sum-k-0n-frac-1kn-choose-k-1-1n-fracn1n/3760847#3760847 – Z Ahmed Jul 25 '20 at 16:10

2 Answers2

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Let $$S=\sum_{r=1}^{3n-1} \frac{(-1)^{r-1} ~r}{3n \choose r}$$ Change $r\to 3n-r$, then $$S=\sum_{r=1}^{3n-1} \frac{(-1)^{3n-r-1}~(3n-r)}{3n \choose 3n-r}= (-1)^{3n}\sum_{s=1}^{3n-1}\frac{(-1)^{r-1}(3n-r)}{{3n \choose s}}$$ $$\implies [1+(-1)^{3n}]S=3n(-1)^{3n-1}\sum_{r=1}^{3n-1}\frac{(-1)^r}{{3n \choose r}}=3n(-1)^{3n-1}\left( \sum_{r=0}^{3n} \frac{(-1)^r}{{3n \choose r}}-(1+(-1)^{3n})\right)$$ Next using Proving that $\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.$ We get $$[1+(-1)^{3n}] S=3n(-1)^{3n-1} [1+(-1)^{3n}]\frac{3n+1}{3n+2}+3n[1+(-1)^{3n}]$$ If $3n$ is even, then $$S=3n \left(1-\frac{3n+1}{3n+2}\right)=\frac{3n}{3n+2},~\text{only if}~n ~\text{is even}$$

For odd $n$ the sum is (perhaps) not doable by hand in a closed form. One may check that $S(n=1)=-1/3, S(n=3)=-7/9.$

Z Ahmed
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\begin{aligned} &\int\limits_0^\infty \frac{x^k}{(1 + x)^{n + 2}} \, dx = \frac{1}{\Gamma(n + 2)} \int\limits_0^\infty x^k \frac{\Gamma(n + 2)}{(1 + x)^{n + 2}} \, dx \\ &= \frac{1}{\Gamma(n + 2)} \int\limits_0^\infty x^k \int\limits_0^\infty y^{n + 1} e^{-(1 + x)y} \, dy \, dx \\ &= \frac{1}{(n + 1)!} \int\limits_0^\infty y^{n + 1} e^{-y} \int\limits_0^\infty x^k e^{-xy} \, dx \, dy \\ &= \frac{1}{(n + 1)!} \int\limits_0^\infty y^{n + 1} e^{-y} \frac{\Gamma(k + 1)}{y^{k + 1}} \, dy \\ &= \frac{k!}{(n + 1)!} \int\limits_0^\infty y^{n - k} e^{-y} \, dy \\ &= \frac{k!}{(n + 1) \cdot n!} \Gamma(n - k + 1) \\ &= \frac{k!(n - k)!}{(n + 1) \cdot n!} \\ &\Rightarrow \boxed{\frac{1}{\binom{n}{k}} = (n + 1) \cdot \int\limits_0^\infty \frac{x^k}{(1 + x)^{n + 2}} \, dx} \end{aligned}

Then

\begin{aligned} S &= \sum_{r = 1}^{2n - 1} \frac{(-1)^{r - 1}r}{\binom{2n}{r}} \\ &= (2n + 1) \sum_{r = 1}^{2n - 1} (-1)^{r - 1}r \int\limits_0^\infty \frac{x^r}{(1 + x)^{2n + 2}} \, dx \\ &= (2n + 1) \int\limits_0^\infty \frac{1}{(1 + x)^{2n + 2}} \left(\sum_{r = 1}^{2n - 1} r(-1)^{r - 1}x^r\right) \, dx \end{aligned}

But

\begin{aligned} \sum_{r = 1}^{2n - 1} r(-1)^{r - 1}x^r &= -x \sum_{r = 1}^{2n - 1} (-1)^rx^{r'} \\ &= -x\left(\sum_{r = 1}^{2n - 1} (-x)^r\right)' \\ &= -x\left(\left(-x \cdot \frac{1 - (-x)^{2n - 1}}{1 - (-x)}\right)'\right) \\ &= x \cdot \left(x \cdot \frac{1 + x^{2n - 1}}{1 + x}\right)' \end{aligned}

Finally

\begin{aligned} S &= (2n + 1) \int\limits_0^\infty \frac{x}{(1 + x)^{2n + 2}} \left(\left(x \cdot \frac{1 + x^{2n - 1}}{1 + x}\right)'\right) \, dx \\ &= \frac{n}{n + 1} \end{aligned}

The calculations are easy but require a lot of operations. The relationship was used $$\boxed{\int\limits_0^\infty x^a e^{-bx} \, dx = \frac{\Gamma(a + 1)}{b^{a + 1}}}.$$

Martin.s
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