Determine whether series is convergent or divergent $\sum_{n=1}^{\infty}\frac{1}{n^2+4}$ I still haven't gotten the hang of how to solve these problems, but when I first saw this one I thought partial fraction or limit.  So I went with taking the limit but the solution manual shows them using the integral test.  
Was I wrong to just take the limit?
$$\sum_{n=1}^{\infty}\frac{1}{n^2+4}$$
Next:
$$\lim_{n\to\infty}\frac{1}{n^2+4}=0$$
So converges by the test for divergence?

Another way,
that needs fewer theorems:
$n^2+4
> n(n-1)$
so
$\frac1{n^2+4}
< \frac1{n(n-1)}
= \frac1{n-1}-\frac1{n}
$.
Therefore,
for any $m > 0$
$\sum_{n=2}^m \frac1{n^2+4}
< \sum_{n=2}^m (\frac1{n-1}-\frac1{n})
= 1-\frac1{m}
< 1
$.
Therefore
$\sum_{n=2}^m \frac1{n^2+4}$
converges as $m \to \infty$.
