Question

Find if the series converges and if so, find its sum

Answer 1

\[\sum_{k=1}^{\infty} 1/(k+2)(k+3)\]

Answer 2

My textbook gives an answer of 1/3 but I can't get that. I split it into partial fractions then do the integral test and get a divergent result

Answer 3

\[\sum_{k=1}^\infty\frac{1}{(k+2)(k+3)}\le\sum_{k=1}^\infty\frac{1}{k^2}\]so the series converges, as you already know.

To find the value, like you said, break up into partial fractions:
\[\frac{1}{(k+2)(k+3)}=\frac{1}{k+2}-\frac{1}{k+3}\]so you have
\[\sum_{k=1}^\infty \left(\frac{1}{k+2}-\frac{1}{k+3}\right)=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\cdots\]See a pattern?