$(a) \sum_{n=2}^{\infty} \frac{3}{n^2+2n}$ Determine whether it converges and find sum Determine whether the following are convergent or divergent. If it is
convergent find its sum. Make sure to fully justify all of your work.
$$(a) \sum_{n=2}^\infty \frac{3}{n^2+3n}$$
Solution (My attempt):
Let $f(n) = a_n$
On $[2,\infty), f(x) = \frac 3 {x^2+3x} > 0$
$$f'(x) = 3\left(\frac{-(2x+3)}{(x^2+3x)^2}\right) < 0, \forall x \in [2,\infty)$$
Therefore the hypothesis of the integral test is met.
Consider
$$\lim_{A\to\infty} \int_2^A \frac 3 {(x)(x+3)} \, dx$$
$$\lim_{A\to\infty} \int_{2}^{A} \left( \frac{1}{x} - \frac{1}{x+3} \right)dx \text{ By pfd}$$
$$\lim_{A\to\infty} \left( \ln \right(\frac A {A+3}\left) - \ln \right(\frac 2 5 \left)\right) $$
$$= \ln 1 - \ln \left(\frac{2}{5}\right)$$
Therefore by Integral test the original series converges.
I was never thought how to find the sum of this type of series. The only way I know how to find the sum of a series is if its geometric. Does anyone know how too find the sum? 

To compute the sum observe
\begin{align}
\frac{1}{n(n+2)} = \frac{1}{2n}-\frac{1}{2(n+2)}
\end{align}
which means
\begin{align}
\sum^\infty_{n=2}\frac{3}{n^2+2n} =&\ \frac{3}{2}\sum^\infty_{n=2}\left( \frac{1}{n}-\frac{1}{n+2}\right)\\
=&\ \frac{3}{2}\sum^\infty_{n=2}\left(\frac{1}{n}-\frac{1}{n+1}\right)+\frac{3}{2}\sum^\infty_{n=2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\\
=&\ \frac{1}{2}+\frac{3}{4}.
\end{align}
