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

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- Jan 12th 2012, 08:33 AMBabyMiloSum of series to infinite
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}$

- Jan 12th 2012, 08:51 AMFernandoRevillaRe: 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 AMBabyMiloRe: Sum of series to infinite
- Jan 12th 2012, 08:55 AMFernandoRevillaRe: Sum of series to infinite
- Jan 12th 2012, 09:22 AMBabyMiloRe: Sum of series to infinite
- Jan 12th 2012, 10:00 AMJesterRe: 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.