## Calculus: Early Transcendentals 8th Edition

Convergent with sum $\frac{11}{6}$
$\Sigma \frac{3}{n(n+3)}$ $\frac{3}{n(n+3)}=\frac{A}{n}+{B}{n+3}=\frac{An+3A}{n(n+3)}+\frac{Bn}{n(n+3)}$ $(A+B)n+3A=3$ $A+B=0$ and $3A=3$ $A=1$ and $B=-1$ $\Sigma \frac{3}{n(n+3)}=\Sigma (\frac{1}{n}-\frac{1}{n+3})=(\frac{1}{1}-\frac{1}{4})+(\frac{1}{2}-\frac{1}{5})+(\frac{1}{3}-\frac{1}{6})+(\frac{1}{4}-\frac{1}{7})+...=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}$ Convergent