find the sum of 1/[n(n+3)]
$\displaystyle 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}) = $
$\displaystyle =\frac{1}{3}\cdot (1 + \frac{1}{2} + \frac{1}{3} +\frac{1}{4} - \frac{1}{4} + \frac{1}{5} - \frac{1}{5} + \dots) = \frac{1}{3}\cdot \frac{11}{6} = \frac{11}{18}$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
The term $\displaystyle \frac{11}{6}$ derives from...
$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}$
In general is...
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n\cdot (n+k)} = \frac{1}{k} \cdot \sum_{i=1}^{k} \frac{1}{i}$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$