Question

find the sum of the series. summation on n=1 going to infinity of (1/n(n+3))

Answer 1

\[\sum_{n=1}^{\infty} 1/n(n+3)\]

Answer 2

Break it up into partial fractions:
\[\frac{1}{n(n+3)}=\frac{A}{n}+\frac{B}{n+3}\]
Solve for A and B, then you have a series like this:
\[\sum_{n=1}^\infty\left(\frac{A}{n}+\frac{B}{n+3}\right)\]