Sum of series to infinite

**BabyMilo** · January 12th 2012, 08:33 AM

$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}$

**FernandoRevilla** · January 12th 2012, 08:51 AM

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

**BabyMilo** · January 12th 2012, 08:53 AM

Originally Posted by FernandoRevilla

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

More detail, please?

**FernandoRevilla** · January 12th 2012, 08:55 AM

Originally Posted by BabyMilo

More detail, please?

What methods have you covered, please?

**BabyMilo** · January 12th 2012, 09:22 AM

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.

**Danny** · January 12th 2012, 10:00 AM

Note that

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

Write some terms in the series

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

and see what happens.

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