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