<s>[INST] Find if $\sum\limits^{\infty}_{n=2} \frac{3}{n^2+3n}$ converges or not. If converges, find its sum. Find if $$\sum^{\infty}_{n=2} \dfrac{3}{n^2+3n}$$
converges or not. If converges, find the sum.
I used the integral test to determine whether the series is convergent or not.
I found $$3\int^{\infty}_{2} \dfrac{1}{x^2+3x}\text{ dx} = \ln(5) - \ln(2)$$
and thus I can conclude that the series converges. But my question asks if it converges, find the sum. How? I evaluated the series at some points, and here is what I get:
$$\dfrac{3}{10} + \dfrac{3}{18} + \dfrac{3}{28}+\dfrac{3}{40} \dots$$
I don't see a pattern or anything, so how can I find the sum?
 [/INST] Observe
\begin{align}
\frac{3}{n(n+3)} = \frac{1}{n}-\frac{1}{n+3}
\end{align}
which means
\begin{align}
\sum^\infty_{n=2}\left(\frac{1}{n}-\frac{1}{n+3}\right) =&\  \left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\ldots\\
=&\ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}.
\end{align}
</s>