evaluate the series $\displaystyle \sum_{k\,=\,2}^{\infty }{\frac{1}{(k+1)(k+2)(k+3)}}$ as many ways as possible

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- Nov 21st 2008, 02:06 PM #1

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- Nov 21st 2008, 02:22 PM #2
$\displaystyle \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$

So $\displaystyle \iiint\frac{dx}{1-x}\bigg|_{x=1}=\sum_{n=0}^{\infty}\frac{1}{(n+1)(n +2)(n+3)}$

EDIT: Sorry, now I have time to finish.

So we know that $\displaystyle \begin{aligned}\iiint\frac{dx}{1-x}&=\frac{-1}{2}\ln(1-x^2)(1-x)^2+\frac{1}{4}-\frac{3}{2}x+\frac{3}{4}x^2\\

&=\iiint\sum_{n=0}^{\infty}x^n\\

&=\sum_{n=0}^{\infty}\frac{x^{n+3}}{(n+1)(n+2)(n+3 )}~\forall{x}\in[-1,1]

\end{aligned}$

So

$\displaystyle \begin{aligned}\bigg[\frac{-1}{2}\ln(1-x^2)(1-x)^2+\frac{1}{4}-\frac{3}{2}x+\frac{3}{4}x^2\bigg]\bigg|_{x=1}&=\sum_{n=0}^{\infty}\frac{1}{(n+1)(n+ 2)(n+3)}\\

&=\frac{1}{4}

\end{aligned}$

So finally $\displaystyle \begin{aligned}\sum_{n=2}^{\infty}\frac{1}{(n+1)(n +2)(n+3)}&=\sum_{n=0}^{\infty}\frac{1}{(n+1)(n+2)( n+3)}-\frac{1}{6}-\frac{1}{24}\\

&=\frac{1}{24}\quad\blacksquare

\end{aligned}$

- Nov 21st 2008, 02:35 PM #3
Here's my approach:

Your series equals $\displaystyle \int_{0}^{1}{\int_{0}^{1}{\left( y\sum\limits_{k=2}^{\infty }{\frac{(xy)^{k+1}}{k+1}} \right)\,dx}\,dy},$ and after some simple calculations this is $\displaystyle -\left( \int_{0}^{1}{\frac{y^{3}+3y^{2}-6y}{6}\,dy}+\int_{0}^{1}{(y-1)\ln (1-y)\,dy} \right).$ First integral is easy to solve, so for the second one I'll perform a double integration solution, hence the series equals $\displaystyle \frac{7}{24}+\int_{0}^{1}{\int_{0}^{y}{\frac{y-1}{1-z}\,dz}\,dy}=\frac{7}{24}-\frac{1}{4}=\frac{1}{24}.$

- Nov 21st 2008, 07:31 PM #4
Ill just go ahead and be the one to suggest the ugliness that is PFD followed by Telescoping Series for this.

$\displaystyle \sum_{n=2}^{\infty}\left\{\frac{1}{2(n+3)}-\frac{1}{n+2}+\frac{1}{2(n+1)}\right\}$

If anyone would like to do it I would be interested to see how many terms you would have to go through to see the actual sum show its face.

- Nov 21st 2008, 07:48 PM #5

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both methods given by

**Mathstud28**and**Krizalid**are good but they're not strong enough to handle the general form, i.e. evaluating: $\displaystyle \sigma_n=\sum_{k=0}^{\infty} \frac{1}{(k+1)(k+2) \cdots (k+n)},$ where $\displaystyle n \geq 2$ is given.

note that the sum here starts from $\displaystyle k = 0,$ which is more usual than k = 2. evaluating $\displaystyle \sigma_n$ is quite easy if you're familiar with gamma and beta functions. here's the solution: first observe that:

$\displaystyle \frac{1}{(k+1)(k+2) \cdots (k+n)}=\frac{k!(n-1)!}{(n-1)!(k+n)!}=\frac{\Gamma(k+1)\Gamma(n)}{(n-1)!\Gamma(k+n+1)}=\frac{B(k+1,n)}{(n-1)!}.$ from here things are straightforward:

$\displaystyle (n-1)!\sigma_n=\sum_{k=0}^{\infty}B(k+1,n)=\sum_{k=0} ^{\infty}\int_0^1t^k(1-t)^{n-1} \ dt=\int_0^1[(1-t)^{n-1}\sum_{k=0}^{\infty}t^k] \ dt$

$\displaystyle =\int_0^1\frac{(1-t)^{n-1}}{1-t} \ dt = \int_0^1 (1-t)^{n-2} \ dt = \frac{1}{n-1} \Longrightarrow \sigma_n=\frac{1}{(n-1)(n-1)!}. \ \ \Box$

as a result, if you really want your sum to start from $\displaystyle k = 2,$ you will get: $\displaystyle \sum_{k=2}^{\infty} \frac{1}{(k+1)(k+2) \cdots (k+n)}=\sigma_n - \frac{1}{n!} - \frac{1}{(n+1)!}=\frac{2}{(n-1)(n+1)!}.$

- Nov 21st 2008, 07:54 PM #6

- Nov 21st 2008, 08:27 PM #7

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right! you can rewrite it more clearly: $\displaystyle \int_0^1\int_0^{x_n} \cdots \int_0^{x_3} \int_0^{x_2} \frac{\ dx_1 \ dx_2 \ \cdots \ dx_n}{1-x_1} =\frac{1}{(n-1)(n-1)!}, \ \ 0 < x_j < 1.$

the challenge now is to prove this result directly, i.e. without using the series!

- Nov 21st 2008, 08:56 PM #8