Simplifying $\frac{^nC_{1}}{1 \times 2}-\frac{^nC_{2}}{2 \times 3}+..+(-1)^{n+1}\frac{^nC_{n}}{n \times (n+1)}$

Question

I was thinking about the following sum -

$\frac{^nC_{1}}{1 \times 2}-\frac{^nC_{2}}{2 \times 3}+\frac{^nC_{3}}{3 \times 4}+..+(-1)^{n+1}\frac{^nC_{n}}{n \times (n+1)} = 1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+..+\frac{1}{n+1}$

How to approach to prove this summation equality?, I could owrite the LHS as $2n - 3n(n-1) + 2n(n-1)(n-2) +...$ after simplification.

Answer 1

Firstly, $$\frac{^nC_{1}}{1 \times 2}+\frac{^nC_{2}}{2 \times 3}+\frac{^nC_{3}}{3 \times 4}+..+\frac{^nC_{n}}{n \times (n+1)} =\left[\frac{^nC_{1}}{1}+\frac{^nC_{2}}{2}+\frac{^nC_{3}}{3}+..+\frac{^nC_{n}}{n}\right]-\left[\frac{^nC_{1}}{2}+\frac{^nC_{2}}{3}+\frac{^nC_{3}}{4}+..+\frac{^nC_{n}}{(n+1)}\right] $$ Now consider, \begin{align*} (1+x)^n-1 & = \binom{n}{1}x+\binom{n}{2}x^2+\dotsb+\binom{n}{n}x^n\\ \int_0^1((1+x)^n-1)\,dx & = \binom{n}{1}\frac{1}{2}+\binom{n}{2}\frac{1}{3}+\dotsb+\binom{n}{n}\frac{1}{n+1}\\ \frac{(1+x)^n-1}{x} & = \binom{n}{1}1+\binom{n}{2}x+\dotsb+\binom{n}{n}x^{n-1}\\ \int_0^1\frac{(1+x)^n-1}{x}\,dx & = \binom{n}{1}\frac{1}{1}+\binom{n}{2}\frac{1}{2}+\dotsb+\binom{n}{n}\frac{1}{n}. \end{align*} So your given series is $$\int_0^1\frac{(1+x)^n-1}{x}\,dx-\int_0^1((1+x)^n-1)\,dx$$

Since the question was edited later:

Since you made the changes in the sign of the actual expression, so you need to redo what I have proposed with $(1-x)^n$.

Answer 2

We can prove the following involving harmonic numbers:

$$\sum_{r=1}^n \frac{(-1)^{r+1}}{r(r+1)} {n\choose r} = H_{n+1} - 1.$$

Start by writing for the LHS

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

Working with the sum term and introducing

$$f(z) = (-1)^{n+1} \frac{(n+1)!}{z-1} \prod_{q=0}^{n+1} \frac{1}{z-q}$$

we have for $2\le r\le n+1$ that

$$\sum_{r=2}^{n+1} \mathrm{Res}_{z=r} f(z) = \sum_{r=2}^{n+1} (-1)^{n+1} \frac{(n+1)!}{r-1} \prod_{q=0}^{r-1} \frac{1}{r-q} \prod_{q=r+1}^{n+1} \frac{1}{r-q} \\ = \sum_{r=2}^{n+1} (-1)^{n+1} \frac{(n+1)!}{r-1} \frac{1}{r!} \frac{(-1)^{n-r+1}}{(n+1-r)!} = \sum_{r=2}^{n+1} \frac{(-1)^{r}}{r-1} {n+1\choose r},$$

so this is the desired sum. Now residues sum to zero and the residue at infinity is zero since $\lim_{R\to\infty} 2\pi R / R / R^{n+2} = 0.$ The remaining poles are at $z=0$ and $z=1.$ We write

$$f(z) = (-1)^{n+1} (n+1)! \frac{1}{z} \frac{1}{(z-1)^2} \prod_{q=2}^{n+1} \frac{1}{z-q}$$

to get

$$\mathrm{Res}_{z=0} f(z) = (-1)^{n+1} (n+1)! \frac{(-1)^n}{(n+1)!} = -1$$

as well as

$$\mathrm{Res}_{z=1} f(z) = (-1)^{n+1} (n+1)! \left. \left(-\frac{1}{z^2} \prod_{q=2}^{n+1} \frac{1}{z-q} - \frac{1}{z} \prod_{q=2}^{n+1} \frac{1}{z-q} \sum_{q=2}^{n+1} \frac{1}{z-q} \right) \right|_{z=1} \\ = (-1)^{n+1} (n+1)! \left(- \frac{(-1)^n}{n!} + \frac{(-1)^n}{n!} H_n \right) = n+1 - (n+1) H_n.$$

Restoring the multiplier in front we thus obtain

$$\frac{1}{n+1} (- \mathrm{Res}_{z=0} f(z) - \mathrm{Res}_{z=1} f(z) ) = \frac{1}{n+1} (1 + (n+1) H_n - (n+1)) \\ = H_n - \frac{n}{n+1} = H_{n+1} - 1.$$

Answer 3

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

$\ds{\sum_{r = 1}^{n}{\pars{-1}^{r + 1} \over r\pars{r + 1}} {n \choose r} = H_{n + 1} - 1:\ {\LARGE ?}}$.

\begin{align} &\bbox[10px,#ffd]{\sum_{r = 1}^{n}{\pars{-1}^{r + 1} \over r\pars{r + 1}} {n \choose r}} = -\sum_{r = 1}^{n} {n \choose r}\pars{-1}^{r}\pars{\int_{0}^{1}x^{r - 1}\,\dd x} \pars{\int_{0}^{1}y^{r}\,\dd y} \\[5mm] = &\ -\int_{0}^{1}\int_{0}^{1}\sum_{r = 1}^{n}{n \choose r}\pars{-xy}^{r}\, {\dd x \over x}\,\dd y = -\int_{0}^{1}\int_{0}^{1}\bracks{\pars{1 - xy}^{n} - 1}\,{\dd x \over x}\,\dd y \\[5mm] = &\ -\int_{0}^{1}\int_{0}^{y}\bracks{\pars{1 - x}^{n} - 1} \,{\dd x \over x}\,\dd y = -\int_{0}^{1}{\pars{1 - x}^{n} - 1 \over x}\int_{x}^{1}\dd y\,\dd x \\[5mm] = &\ -\int_{0}^{1}{\pars{1 - x}^{n + 1} - 1 + x \over x}\,\dd x = \int_{0}^{1}x\,{x^{n} - 1 \over x - 1}\,\dd x = \int_{0}^{1}\sum_{k = 1}^{n}x^{k}\,\dd x \\[5mm] = &\ \sum_{k = 1}^{n}{1 \over k + 1} = \sum_{k = 2}^{n + 1}{1 \over k} = \bbx{H_{n + 1} - 1} \\ & \end{align}