Determine whether the follwing series is convergent or divergent. If it is convergent, find the sum.
$$\sum\limits_{n=1}^\infty\frac{1}{n^2+5n+6}$$
Here's my work so far:
$\lim_\limits{n\to\infty}\frac{1}{n^2+5n+6} = 0$
$\therefore\;$ the series is convergent.
I don't think it's a geometric series since there is no common factor between the consecutive terms. Because it isn't a geometric series, I'm at a loss as to what formula to use.
Best Answer
\begin{align}\sum_{n=1}^\infty\frac1{n^2+5n+6}&=\sum_{n=1}^\infty\frac1{(n+2)(n+3)}\\&=\sum_{n=1}^\infty\left(\frac1{n+2}-\frac1{n+3}\right)\\&=\left(\frac13-\frac14\right)+\left(\frac14-\frac15\right)+\left(\frac15-\frac16\right)+\cdots\\&=\frac13-\lim_{n\to\infty}\frac1{n+2}\\&=\frac13.\end{align}