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?
 A: We only have the following statement to be true:
$$\text{If $\sum_{n=1}^{\infty} a_n$ converges, then $a_n \to 0$.}$$ The converse of the above statement is not true, i.e.,
$$\text{if $a_n \to 0$, then $\displaystyle \sum_{n=1}^{\infty} a_n$ converges is an incorrect statement.}$$
For instance, $\displaystyle \sum_{n=1}^{\infty} \dfrac1n$ diverges, even though $\dfrac1n \to 0$.
To prove your statement, note that $\dfrac1{n^2+4} < \dfrac1{n^2}$ and make use of the fact that $\displaystyle \sum_{n=1}^{\infty} \dfrac1{n^2}$ converges to conclude that $\displaystyle \sum_{n=1}^{\infty}\dfrac1{n^2+4}$ converges.
A: Just expounding on $\frac{1}{n^2 + 4} < \frac{1}{n^2}$ in the above answer. 
This holds using 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. 
A: 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 $m \to \infty$.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\sum_{n = 1}^{\infty}{1 \over n^{2} + 4}
&=
\sum_{n = 0}^{\infty}{1 \over \pars{n + 1 + 2\ic}\pars{n + 1 - 2\ic}}
=
{\Psi\pars{1 + 2\ic} - \Psi\pars{1 - 2\ic} \over \pars{1 + 2\ic} - \pars{1 - 2\ic}}
=
{1 \over 2}\,\Im\Psi\pars{1 + 2\ic}
\\[3mm]&=
{1 \over 2}\bracks{-\,{1 \over 4} + {1 \over 2}\,\pi\coth\pars{2\pi}}
\end{align}
$$
\color{#0000ff}{\large\sum_{n = 1}^{\infty}{1 \over n^{2} + 4}}
=
\color{#0000ff}{\large{1 \over 4}\bracks{\pi\coth\pars{2\pi} - {1 \over 2}}}
\approx 0.6604
$$
