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

Question

Deduce that $\displaystyle {n \choose 0}\dfrac{1}{1\cdot2}-{n \choose 1}\dfrac{1}{2\cdot3}+{n \choose 2}\dfrac{1}{3\cdot4}+...(-1)^n{n \choose n}\dfrac{1}{(n+1)\cdot(n+2)}=\dfrac{1}{n+2}$

We know $\dfrac{1}{n+2}=\displaystyle \int_{0}^{1}t^{n+2-1}dt$

Now $\displaystyle {n \choose 0}\dfrac{1}{1\cdot2}-{n \choose 1}\dfrac{1}{2\cdot3}+{n \choose 2}\dfrac{1}{3\cdot4}+...(-1)^n{n \choose n}\dfrac{1}{(n+1)\cdot(n+2)}=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)(k+2)}\bigg)=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)}-\dfrac{1}{k+2}\bigg)$

$=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)}\bigg)-\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+2)}\bigg)$

$=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\int_{0}^1 t^{k+1-1}dt \space - \space \displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k \int_{0}^1 t^{k+2-1}dt$

$=\displaystyle \int_{0}^1 t^{1-1}\bigg(\sum_{k=0}^{n}{n\choose k}(-t)^k\bigg) dt- \int_{0}^1 t^{2-1}\bigg(\sum_{k=0}^{n}{n\choose k}(-t)^k\bigg) dt$

$=\displaystyle \int_{0}^1 t^{1-1}(1-t)^{n}dt \space - \space \int_{0}^1 t^{2-1}(1-t)^{n}dt$

$=\displaystyle \large \beta(1,n+1)$-$\displaystyle \large\beta(2,n+1)$

$=\dfrac{\Gamma(1)\Gamma(n+1)}{\Gamma(1+n+1)}-\dfrac{\Gamma(2)\Gamma(n+1)}{\Gamma(2+n+1)}= \underbrace{\dfrac{1}{n+2}}_{\text{which is exactly what I proved}}$

PS @Chappers Thankyou all users for correcting one nasty mistake.

Answer 1

Your integral is incorrect: for $a>0$, $$ \frac{1}{a} = \int_0^1 t^{a-1} \, dt, $$ so the integrals should be $$ \frac{1}{k+1} = \int_0^1 t^{k} \, dt $$ and $$ \frac{1}{k+2} = \int_0^1 t^{k+1} \, dt, $$ and then you get $$ B(1,n+1)-B(2,n+1) = \frac{1}{n+2} $$ as expected.

Answer 2

Hint:

Multiply and divide by $(n+1)(n+2)$ to turn the question into $$\frac {1}{(n+1)(n+2)} \sum_{k=0}^n (-1)^k \binom {n+2}{k+2}= \frac {1}{(n+1)(n+2)}\left[ \left(\sum_{k=-2}^n (-1)^k \binom {n+2}{k+2}\right)+(n+2)-1\right]=\frac {1}{n+2} $$

Answer 3

Alternatively $$\dfrac{\binom nk}{(k+1)(k+2)}=\dfrac{\binom{n+2}{k+2}}{(n+1)(n+2)}$$

Now set $a=b=1$ in $$(a-b)^{n+2}=?$$

Answer 4

$$\frac{1}{(k+1)(k+2)}=\frac{1}{k+1}-\frac{1}{k+2} = \int_{0}^{1} x^k (1-x)\,dx $$ hence $$\begin{eqnarray*} \sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{(k+1)(k+2)} &=& \int_{0}^{1}(1-x)\sum_{k=0}^{n}\binom{n}{k}(-x)^k\,dx\\ &=& \int_{0}^{1}(1-x)^{n+1}\,dx = \int_{0}^{1} x^{n+1}\,dx = \frac{1}{n+2}.\end{eqnarray*}$$

Answer 5

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