## Calculus 8th Edition

$$\frac{11}{18}$$
$$\Sigma_{n=1}^{\infty}\frac{1}{n(n+3)}=\Sigma_{n=1}^{\infty}\frac{1}{3}(\frac{1}{n}-\frac{1}{n+3})$$ $$\Sigma_{n=1}^{\infty}\frac{1}{3}(\frac{1}{n}-\frac{1}{n+3})=\frac{1}{3}\Sigma_{n=1}^{N}\frac{1}{n}-\frac{1}{3}\Sigma_{n=1}^{N}\frac{1}{n+3}$$ $$=\frac{1}{3}(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{N+1}-\frac{1}{N+2}-\frac{1}{N+3}+....)$$ Note that $$\lim\limits_{N \to \infty}=\frac{1}{3}(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{N+1}-\frac{1}{N+2}-\frac{1}{N+3}+..)=\frac{1}{3}(1+\frac{1}{2}+\frac{1}{3})$$ $$=\frac{11}{18}$$