TITLE: What steps should I be doing to determine if this series is convergent or divergent?
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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.

REPLY [4 votes]: $$n^2 + 3n > n^2 \implies \frac{1}{n^2 +3n} < \frac{1}{n^2}$$
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.