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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.

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    $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ Commented Apr 29, 2018 at 18:41
  • $\begingroup$ Why do you think there is an error? You have arrived to $\frac{1}{n+2}$ which is correct. $\endgroup$
    – Sil
    Commented Apr 29, 2018 at 18:53
  • $\begingroup$ @Sil I corrected it now. Thanks $\endgroup$
    – Saradamani
    Commented Apr 29, 2018 at 18:54
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    $\begingroup$ Note you don't need to let people know the question is answered: the site does this by displaying the number of answers and by marking the post green when you have accepted one. $\endgroup$
    – Pedro
    Commented Apr 29, 2018 at 19:00
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    $\begingroup$ I was refering to the content of the question. Originally you had the wrong proof and you asked where the issue was. Then after you got an answer, you changed the question so that there is no longer an issue in the proof. But that renders the question non sensical (and change in question title will not really solve that). I am not sure that using SE as a place for notes is a good idea (for example it might get deleted, you might want to check something like overleaf.com). But even then, there are ways to update the question without changing its meaning. $\endgroup$
    – Sil
    Commented Apr 29, 2018 at 19:15

5 Answers 5

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

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    $\begingroup$ thanks but I cannot understand where I was wrong! Why was it giving something else?? $\endgroup$
    – Saradamani
    Commented Apr 29, 2018 at 18:37
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    $\begingroup$ @Saradamani: because you wrote that $\frac{1}{k+1}=\int_{0}^{1}x^{k+1}\,dx.$ $\endgroup$ Commented Apr 29, 2018 at 18:43
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    $\begingroup$ Bad Bad Bad me! I corrected it thanks! $\endgroup$
    – Saradamani
    Commented Apr 29, 2018 at 18:44
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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.

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  • $\begingroup$ (+1). Finally an answer that actually answers OP's question. $\endgroup$
    – timon92
    Commented Apr 29, 2018 at 18:39
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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Hint:

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

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

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

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