# Sum of series to infinite

• Jan 12th 2012, 08:33 AM
BabyMilo
Sum of series to infinite
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}$
• Jan 12th 2012, 08:51 AM
FernandoRevilla
Re: Sum of series to infinite
$\displaystyle \frac{u_{n+1}}{u_n}=\ldots=\frac{n}{n+3}=\frac{ \alpha n+\beta}{\alpha n+\gamma}$ (hipergeometric series) . According to a well known theorem , $\displaystyle S=\frac{u_1\gamma}{\gamma-(\alpha+\beta)}=\ldots=\frac{1}{4}$
• Jan 12th 2012, 08:53 AM
BabyMilo
Re: Sum of series to infinite
Quote:

Originally Posted by FernandoRevilla
$\displaystyle \frac{u_{n+1}}{u_n}=\ldots=\frac{n}{n+3}=\frac{ \alpha n+\beta}{\alpha n+\gamma}$ (hipergeometric series) . According to a well known theorem , $\displaystyle S=\frac{u_1\gamma}{\gamma-(\alpha+\beta)}=\ldots=\frac{1}{4}$

• Jan 12th 2012, 08:55 AM
FernandoRevilla
Re: Sum of series to infinite
Quote:

Originally Posted by BabyMilo

What methods have you covered, please?
• Jan 12th 2012, 09:22 AM
BabyMilo
Re: Sum of series to infinite
Quote:

Originally Posted by FernandoRevilla
What methods have you covered, please?

I havent been taught how to do this really.
I have covered the test for convergent but the book asks for this.
I used partial fraction which they all diverges.
• Jan 12th 2012, 10:00 AM
Jester
Re: Sum of series to infinite
Note that

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

Write some terms in the series

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

and see what happens.