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

Various examinations ask students to prove that $$\frac{1}{{n \choose k}}=(n+1) \int_{0}^{1} x^k (1-x)^{n-k} dx ~~~~~(1)$$ by evaluating the integral $$\int_{0}^{1} (tx+1-x)^n~dt$$ two ways. When I came across (1), I could prove that $$\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.~~~~(2)$$ as below: $$S_n=\frac{1}{n+1}\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=\int_{0}^1 \sum_{k=0}^{n} (-1)^k x^k (1-x)^{n-k} dx = \int_{0}^{1} (1-x)^n \sum_{k=0}^{n}\left(\frac{-x}{1-x} \right)^k dx =\int_{0}^{1}(1-x)^n \frac{ \left( \frac{-x}{1-x}\right)^{n+1}-1}{\left(\frac{-x}{1-x}\right)-1}dx$$ $$=\int_{0}^{1}[(1-x)^{n+1}-(-x)^{n+1}] dx =\frac{1+(-1)^n}{n+2}.~~~~(3)$$ Therefore, $$S_{2m+1}=0 ~~~~(4) ~~~ \mbox{and}~~~ S_{2m}= \frac{2m+1}{m+1}.~~~~(5)$$ Now the question is: Can one prove (2) in some other way(s)?

• Other solutions are here math.stackexchange.com/q/1402886 and here math.stackexchange.com/q/2027749 Jun 5, 2019 at 8:22
• Wow! wonderful solutions in 2015 and 2016, but while posting this question I got actually dissimilar questions in the listed similar questions! Jun 5, 2019 at 8:37

A telescoping approach:

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}}} &=\sum_{k=0}^n(-1)^k\frac{k!(n-k)!}{n!}\\ &=\sum_{k=0}^n(-1)^k\frac{k!(n-k)!}{n!}\cdot\frac{(n-(k-1))+(k+1)}{n+2}\\ &=\frac{n+1}{n+2}\sum_{k=0}^n(-1)^k\frac{k!(n-(k-1))!+(k+1)!(n-k)!}{(n+1)!}\\ &=\frac{n+1}{n+2}\sum_{k=0}^n\left((-1)^k\frac{1}{\binom{n+1}{k}}-(-1)^{k+1}\frac{1}{\binom{n+1}{k+1}}\right)\\ &\,\,\color{blue}{=\frac{n+1}{n+2}\left(1+(-1)^n\right)} \end{align*}

and the claim follows.

The general formula is:

$$\sum\limits_{k=0}^n\frac{(-1)^k}{\binom x k} = \left(1+\frac{(-1)^n}{\binom {x+1} {n+1}}\right)\frac{x+1}{x+2}$$

Value range: $$\enspace n\in\mathbb{N}_0~,~~ x\in\mathbb{C}\setminus\{n-k|k\in\mathbb{N}\}$$

The proof by induction with respect to $$~n~$$ is based on the following equation: $$\frac{1}{\binom x {n+1}}\frac{x+2}{x+1} = \frac{1}{\binom {x+1} {n+1}} + \frac{1}{\binom {x+1} {n+2}}$$

• $n$ in my question is normal. Congratulations, you have proved a more general result where $x \ge n.$ Jun 5, 2019 at 17:33
• @DrZafarAhmedDSc : Thanks for your kind words and for the hint with the value range. I’ve added. ;) Jun 6, 2019 at 7:59
• excellent the extension to complex values Jun 6, 2019 at 14:29
• @GCab : Thank you! ;) Jun 6, 2019 at 20:13

I think that the intuition about your result can be the following:

$$f(n,k):=\int_0^1 x^k(1-x)^{n-k}>0$$

and you can observe that

$$1=1^n=((1-x)+x)^n=\sum_{k=0}^n\binom{n}{k}(1-x)^{n-k}x^k$$

so if you integrate the two members with respect to $$x$$ you get that

$$\int_0^11dx=1=$$

$$\int_0^1 (\sum_{k=0}^n\binom{n}{k}(1-x)^{n-k}x^k)dx=$$

$$=\sum_{k=0}^n\binom{n}{k}\int_0^1x^n(1-x)^{n-k}dx$$

so

$$1=\sum_{k=0}^n\binom{n}{k}f(n,k)$$

and you can observe that the identity is satisfied when

$$f(n,k)=\frac{1}{\binom{n}{k}}\frac{1}{n+1}$$

So we want prove by induction on $$k>1$$ for all fixed $$n\geq k$$ that $$f(n,k)=\frac{1}{\binom{n}{k}}\frac{1}{n+1}$$

For $$k=0$$ you have that

$$f(n,0)=\int_0^1x^0(1-x)^{n-0}dx=-\frac{1}{n+1}[(1-x)^{n+1}]|_0^1=\frac{1}{n+1}$$

Now we can hypothesize that the sentence is true for some $$k-1$$ and we want prove that it is true for $$k$$:

$$f(n,k)=\int_0^1x^k(1-x)^{n-k}dx=$$

$$-\frac{1}{n-k+1}\int_0^1x^kD((1-x)^{n-k+1})dx=$$

$$=\frac{1}{n-k+1}k\int_0^1x^{k-1}(1-x)^{n-(k-1)}dx=$$

$$=\frac{k}{n-k+1}f(n,k-1)=$$

$$\frac{1}{\binom{n}{k-1}}\frac{1}{n+1} \frac{k}{n-k+1 }=$$

$$=\frac{(k-1)!k(n-(k-1))!}{n!}\frac{1}{(n+1)(n-(k-1))}=$$

$$= \frac{1}{\binom{n}{k}}\frac{1}{n+1}$$

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ $$\ds{\bbox[15px,#ffd]{\sum_{k = 0}^{n}{\pars{-1}^{k} \over {n\choose k}} = \bracks{1 + \pars{-1}^{n}}{n + 1 \over n + 2}}:\ {\Large ?}}$$

\begin{align} \sum_{k = 0}^{n}{\pars{-1}^{k} \over {n\choose k}} & = \sum_{k = 0}^{n}\pars{-1}^{k}\,{k!\,\pars{n - k}! \over n!} \\[5mm] & = \pars{n + 1}\sum_{k = 0}^{n}\pars{-1}^{k}\, {\Gamma\pars{k + 1}\Gamma\pars{n - k + 1} \over \Gamma\pars{n + 2}} \\[5mm] & = \pars{n + 1}\sum_{k = 0}^{n}\pars{-1}^{k}\, \int_{0}^{1}t^{k}\pars{1 - t}^{n - k}\,\dd t \\[5mm] & = \pars{n + 1}\int_{0}^{1}\pars{1 - t}^{n} \sum_{k = 0}^{n}\pars{-\,{t \over 1 - t}}^{k}\,\dd t \\[5mm] & = \pars{n + 1}\int_{0}^{1}\pars{1 - t}^{n}\, {\bracks{-t/\pars{1 - t}}^{n + 1} - 1 \over -t/\pars{1 - t} - 1}\,\dd t \\[5mm] & = \pars{n + 1}\int_{0}^{1}\pars{1 - t}^{n + 1}\, \bracks{1 - \pars{-\,{t \over 1 - t}}^{n + 1}}\,\dd t \\[5mm] & = \pars{n + 1}\int_{0}^{1} \bracks{\pars{1 - t}^{n + 1} + \pars{-1}^{n}\, t^{n + 1}}\,\dd t \\[5mm] & = \pars{n + 1} \bracks{-\,{\pars{1 - t}^{n + 2} \over n + 2} + \pars{-1}^{n}\, {t^{n + 2} \over n + 2}}_{\ 0}^{\ 1} \\[5mm] & = \pars{n + 1} \bracks{{\pars{-1}^{n} \over n + 2} + {1 \over n + 2}} \\[5mm] & = \bbx{\bracks{1 + \pars{-1}^{n}}{n + 1 \over n + 2}} \\ & \end{align}