The problem is: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$
The first thing I did was use the divergence test which didn't help since the result of the limit was 0.
If I multiply it through, the result is $\sum_{n=1}^{\infty} \frac{1}{n^2+3n}$
I'm wondering if I can consider this as a p-series and simply use the largest power. In this case the power would be 2 which would mean it converges. If this is the correct way to go about this, how do I find where it converges to.
Bound it above! Note $n(n+3)=n^2+3n>n^2$
so $\frac{1}{n(n+3)}<\frac{1}{n^2}$
Each term is clearly > 0 btw.
So! $\sum\frac{1}{n(n+3)}<\sum\frac{1}{n^2}$ which you ought to know (but can trivially show) converges.
Finally a question I can answer here!
Note that $\frac1{n(n+3)}=\frac13\left(\frac1n-\frac1{n+3}\right)$ so this is a telescoping sum $$\sum_{n=1}^m \frac1{n(n+3)}=\frac13\left(1+\frac 12+\frac13-\frac1{m+1}-\frac1{m+2}-\frac1{m+3}\right)\to \frac{11}{18}.$$
First, use estimations
$$ n^2 + 3n \geq n^2 \implies \frac{1}{n^2 + 3n } \leq \frac{1}{n^2} $$
Secondly, show that $\sum \frac{1}{n^2}$ converges. In fact, it does. More generally,
$$ \sum \frac{1}{n^p} \; \; \text{converges when} \; \; p > 1 $$
Third, use the comparison theorem: if $a_n \geq b_n $ for all $n$ and $\sum a_n$ converges, then $\sum b_n$ must converge as well (Proof?)
Now, as an application of this theorem, with $a_n = \frac{1}{n^2} $ and $b_n = \frac{1}{n^2 + 3n}$, we notice that your series
$$ \sum \frac{1}{n^2 + 3n} $$
must converge.
$$n^2 + 3n > n^2 \implies \frac{1}{n^2 +3n} < \frac{1}{n^2}$$
Use the Comparison Test which states that if $\sum a_n$ and $\sum b_n$ are such that $0 \le a_n \le b_n$, if $\sum b_n$ converges, then $\sum a_n$ converges.
Since $0 < \sum \frac{1}{n^2 +3n} < \sum \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges, then $\sum \frac{1}{n^2 +3n}$ converges.
Edit. Note: $\sum \frac{1}{n^2}$ converges since it is a p-series $$f(x) = \frac{1}{X^p}$$ with $p > 1$ and hence it converges.
