Determine the convergence of a recursive sequence Determine the convergence of the sequence defined by $\sum_{n \in \mathbb{N}}a_n$ such as
$$a_1=1,\quad a_{n+1}=\frac{2+\cos(n)}{\sqrt{n}}a_n$$

By the test of the general term, we have
$$\frac{1}{\sqrt{n}} \le \frac{2 + \cos{(n)}}{\sqrt{n}} \le \frac{3}{\sqrt{n}}  $$ 
So, by the theorem of the sandwich,
$$ \lim_{n \rightarrow \infty}\frac{1}{\sqrt{n}} 
=\lim_{n \rightarrow \infty}\frac{2 + \cos{(n)}}{\sqrt{n}}
=\lim_{n \rightarrow \infty}\frac{3}{\sqrt{n}} $$
We get, $\lim_{n \rightarrow \infty}\frac{2+\cos(n)}{\sqrt{n}}a_n=0$.

How to a continue my proof?
If we use the ratio test, 
$$
\left| \frac{a_{n+1}}{a_n} \right|
= \left|\frac{2+\cos{(n)}}{\sqrt n} \right| 
\leq \frac{3}{\sqrt{n}} \to 0.
$$
So the series converges absolutely.
For $n\geq 16 $ we have $|a_{n+1}|=(|2+\cos n|/\sqrt n)|a_n|\leq(3/4)|a_n|$ so for $n\geq  16$ we have $|a_{n+1}|\leq (3/4)^{n-15}|a_{16}|.$