<s>[INST] Does $\sum_1^\infty \frac{n}{n^2 + 4}$ converge or diverge? Is my solution correct? Does $\sum_1^\infty \frac{n}{n^2 + 4}$ converge or diverge?
I am confused because my friend insists the series converges conditionally. I think the series diverges. Here is my process and solution:
$\sum_1^\infty \frac{n}{n^2 + 4}$ 
Limit Comparison Test:
Let $\sum_1^\infty \frac{n}{n^2 + 4} = \sum_1^\infty a_n$
Let $\sum_1^\infty b_n = \sum_1^\infty \frac{1}{n}$
Since $$\lim \limits_{n \to \infty} | \frac{a_n}{b_n}| = \lim \limits_{n \to \infty} | \frac{\frac{n}{n^2 + 4}}{\frac{1}{n}}| = \lim \limits_{n \to \infty} \frac{n^2}{n^2 + 4} = \lim \limits_{n \to \infty} \frac{n^2}{n^2(1 + \frac{4}{n^2})} = \lim \limits_{n \to \infty} \frac{1}{1 + \frac{4}{n^2}} = 1$$
and $1 > 0$ and $1$ is a finite number, 
we can say that the behavior of $\sum_1^\infty a_n$ is the same as the behavior of $\sum_1^\infty b_n$.
Since $\sum_1^\infty b_n$ diverges ( $\sum_1^\infty \frac{1}{n}$ diverges by p-series, $p=1$), $\sum_1^\infty a_n$ diverges.
So by the Limit Comparison Test, $\sum_1^\infty \frac{n}{n^2 + 4}$  diverges.
Right?

Also, this problem was an exercise in the Ratio/Root Test section. Both tests, however seem to fail. Can the convergence be solved with the Root Test or the Ratio Test?
 [/INST] Direct comparison is a little easier here:
$$\sum_{n=1}^\infty \frac{n}{n^2+4} \geq \sum_{n=2}^\infty \frac{n}{n^2+4} \geq \sum_{n=2}^\infty \frac{n}{n^2+n^2} = \frac{1}{2}\sum_{n=2}^\infty \frac{1}{n} = \infty$$
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