Convergence or divergence $ \sum \frac{2+\sin n\theta}{n^2 +\sin n \theta}$? (using Cauchy test) 
Determine convergence or divergence
  $$ \sum\limits_{n=1}^{\infty} \frac{2+\sin n\theta}{n^2 +\sin n \theta}$$

Considering $a_n = \frac{2+\sin n\theta}{n^2 +\sin n \theta}$, let's apply Cauchy root test:
Since $ -1 \leq \sin n\theta \leq 1 \implies \frac{2+\sin n\theta}{n^2 +\sin n \theta} \geq 0, \forall n\in[2,\infty)$
$$ \overline\lim_{n \rightarrow \infty} a_n^{1/n} = \left(\frac{3}{n^2+1}\right)^{1/n}$$
Since $\frac{3}{n^2+1}$ is decreasing and when $n=2$:
$$ \left(\frac{3}{2^2+1}\right)^{1/2} \approx 0.77 < 1 $$
We can conclude that 
$$ \overline\lim_{n \rightarrow \infty} a_n^{1/n} = \left(\frac{3}{n^2+1}\right)^{1/n}<1, \forall n \in[2, \infty)$$
By the Cauchy's Root Test, It follows that:  $$ \sum \frac{2+\sin n\theta}{n^2 +\sin n \theta} < \infty$$
First time with this Cauchy test. Not too certain if it is the correct way to deal with "$\overline\lim$". Am I doing this correctly? Is there a more formal approach to show that the $n^{th}$ root is less than $1$?
 A: You have not applied the root test correctly. Although it is quite true that $3/(n^2 + 1)$ is decreasing, that does not imply that the $n$-th root of it is. In fact, 
$$\lim_{n \to \infty} \left(\frac 3 {n^2 + 1}\right)^{1/n} = 1$$
and the root test is not applicable to this series. For a similar case, consider the fact that $1/n$ decreases to zero, but
$$\left(\frac 1 n\right)^{1/n} = e^{\log n / n} \to e^0 = 1$$
in the limit.

Also, there are a couple other issues. First, if you're using this test, you should have $n^2 - 1$ rather than $+1$. Secondly,

$$ \overline\lim_{n \rightarrow \infty} a_n^{1/n} = \left(\frac{3}{n^2+1}\right)^{1/n}<1, \forall n \in[2, \infty)$$

doesn't make sense at all. You can't have $n$ on the right hand side of this equality, because it's got a very specific meaning on the left hand side. The limit cannot depend on $n$, since $n \to \infty$.

For a different approach, just try direct comparison, using the fact that your summands are all smaller than $3/(n^2 - 1)$ for $n > 1$.
A: I would use equivalents and the comparison test:
$$0<\frac{2+\sin n\theta}{n^2+\sin n\theta}\le \frac3{n^2+\sin n\theta}\sim_\infty \frac3{n^2},$$
which is a convergent Riemann series.
A: $$\frac{2+\sin{n\theta}}{n^2+\sin{n\theta}}<\frac{3}{\sqrt{n^3}}$$
