## Multivariable Calculus, 7th Edition

The series is convergent and the sum is $\frac{11}{6}$
$\Sigma^{\infty}_{n=1} \frac{3}{n(n+3)}$ $S_{n} =\Sigma^{n}_{k=1} \frac{3}{k(k+3)}$ $=\frac{3}{1 \times 4}+ \frac{3}{2 \times 5} + ... +\frac{3}{n(n+3)}$ $S_{n} = \Sigma^{n}_{k=1} \frac{3}{k(k+3)}$ $=\Sigma^{n}_{k=1} (\frac{1}{k} - \frac{1}{k+3})$ $=(1-\frac{1}{4}) + (\frac{1}{2}-\frac{1}{5})+(\frac{1}{3}-\frac{1}{6})+(\frac{1}{4}-\frac{1}{7})+...+(\frac{1}{n} - \frac{1}{n+3})$ $=1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+3}$ $\lim\limits_{n \to \infty}S_{n} = \lim\limits_{n \to \infty}(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+3})$ $=1+\frac{1}{2}+\frac{1}{3}-\frac{1}{\infty+3}$ $=1+\frac{1}{2}+\frac{1}{3}-\frac{1}{\infty}$ $=1+\frac{1}{2}+\frac{1}{3}-0$ $=\frac{6+3+2}{6}$ $=\frac{11}{6}$