find the sum of the series question **

**twilightstr** · May 14th 2009, 11:59 PM

find the sum of 1/[n(n+3)]

**chisigma** · May 15th 2009, 12:25 AM

$S= \sum_{n=1}^{\infty} \frac{1}{n\cdot (n+3)} = \frac{1}{3}\cdot \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+3}) =$

$=\frac{1}{3}\cdot (1 + \frac{1}{2} + \frac{1}{3} +\frac{1}{4} - \frac{1}{4} + \frac{1}{5} - \frac{1}{5} + \dots) = \frac{1}{3}\cdot \frac{11}{6} = \frac{11}{18}$

Kind regards

$\chi$ $\sigma$

**twilightstr** · May 15th 2009, 01:28 AM

why multiply by 11/6?

**chisigma** · May 15th 2009, 01:40 AM

The term $\frac{11}{6}$ derives from...

$1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}$

In general is...

$\sum_{n=1}^{\infty} \frac{1}{n\cdot (n+k)} = \frac{1}{k} \cdot \sum_{i=1}^{k} \frac{1}{i}$

Kind regards

$\chi$ $\sigma$

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