6

$$\forall k\ge 1\in\mathbb N\\\displaystyle\sum\limits_{i=1}^\infty\frac1{i(i+1)(i+2)...(i+k)}=?$$

Try$(1)$I tried to apply vieta If it be considered like this ;

$$\frac1{i(i+1)..(i+k)}=\dfrac{A_{0}}{i}+\dfrac{A_{1}}{i+1}+...+\dfrac{A_{k}}{i+k}$$

Try$(2)$

$$\frac1{i(i+1)..(i+k)}=\dfrac{(i-1)!}{(i+k)!}=\dfrac{(i-1)!}{(i+k)!}\dfrac{(k+1)!}{(k+1)!}=\dfrac{1}{(k+1)!}\dfrac{1}{\dbinom{i+k}{i-1}}$$

But I couldn't get any usefull equalition.I want to calculate this series but how?

By the way, I exactly know that this series is convergence.

Micheal Brain Hurts
  • 2,380

4 Answers4

5

Hint. One may observe that $$ \begin{align} \frac{k}{i(i+1)(i+2)...(i+k)}&=\frac{(i+k)-i}{i(i+1)(i+2)...(i+k)} \\\\&=\frac{1}{i(i+1)(i+2)...(i+k-1)}-\frac{1}{(i+1)(i+2)...(i+k)} \end{align} $$ then one may see that terms telescope.

Olivier Oloa
  • 120,989
5

Since $$ \frac1{i(i+1)(i+2)\cdots(i+k)}=\frac1k\left(\frac1{i(i+1)(i+2)\cdots(i+k-1)}-\frac1{(i+1)(i+2)\cdots(i+k)}\right) $$ we can use Telescoping Series to get $$ \sum_{i=1}^\infty\frac1{i(i+1)(i+2)\cdots(i+k)}=\frac1{k\,k!} $$

Abr001am
  • 746
robjohn
  • 345,667
3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[15px,#ffe]{\ds{\left.\sum_{i = 1}^{\infty}{1 \over i\pars{i + 1}\pars{i + 2}\ldots\pars{i + k}}\right\vert_{\ k\ \in\ \mathbb{N}_{\ \geq\ 1}}}} = \sum_{i = 1}^{\infty}{1 \over i^{\,\overline{k + 1}}} = \sum_{i = 1}^{\infty}{1 \over \Gamma\pars{i + k + 1}/\Gamma\pars{i}} \\[5mm] = &\ {1 \over k!}\sum_{i = 1}^{\infty}{\Gamma\pars{i}\Gamma\pars{k + 1} \over \Gamma\pars{i + k + 1}} = {1 \over k!}\sum_{i = 1}^{\infty}\int_{0}^{1}t^{i - 1}\pars{1- t}^{k}\,\dd t = {1 \over k!}\int_{0}^{1}\pars{1- t}^{k - 1}\,\dd t = \bbx{\ds{1 \over k\ k!}} \end{align}

Felix Marin
  • 89,464
0

In the answer I referred to when I voted to close, I showed that $\sum_{j=1}^m \dfrac{1}{\prod_{k=0}^{n} (j+k)} = \dfrac1{n}\left(\dfrac1{n!}-\dfrac1{\prod_{k=0}^{n-1} (m+1+k)}\right) $.

This follows from

$\begin{array}\\ \dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=0}^{n-1} (x+1+k)} &=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=1}^{n} (x+k)}\\ &=\dfrac1{\prod_{k=1}^{n-1} (x+k)}\left(\dfrac1{x}-\dfrac1{x+n}\right)\\ &=\dfrac1{\prod_{k=1}^{n-1} (x+k)}\left(\dfrac{n}{x(x+n)}\right)\\ &=\dfrac{n}{\prod_{k=0}^{n} (x+k)}\\ \end{array} $

marty cohen
  • 107,799