Test for convergence/divergence of $\sum_{n=1}^{\infty}\frac{\sqrt{n}\cos(n)}{n+8}$ Given the series

$$\sum_{n=1}^{\infty}\frac{\sqrt{n}\cos(n)}{n+8}$$

I just came across the following question from an book. I need to test for convergence/divergence. My guess it is convergent, but how to show it?
Any help is appreciated. Thanks.
 A: Let $a_n = \dfrac{\sqrt{n}}{n+8}$, $b_n = \cos n$. Clearly $\exists n_0 \in \mathbb{N}$ such that $a_n$ decreases monotonically $\forall n \geq n_0$, and $\displaystyle{\lim _{n \rightarrow \infty} a_n } = 0$.
Moreover, $\exists M \geq 0 : $$\left |\displaystyle{\sum _{n=n_0} ^ N \cos n} \right |\leq M $ $\forall N \geq n_0$. Therefore, using Dirichlet's test $\displaystyle {\sum _{n=n_0} ^\infty a_n b_n}$ converges, and therefore $\displaystyle {\sum _{n=1} ^\infty a_n b_n}$ converges too.
NOTE:
To prove that $C_N = \displaystyle{ \sum_{n=0} ^{N} \cos(n)}$ is bounded, let's consider $S_N = \displaystyle{ \sum_{n=0} ^{N} e^{in}}$. Then, $C_N = \Re (S_N)$, so we just have to prove that $|S_N|$ is bounded.
$$S_n = \displaystyle{ \sum_{n=0} ^{N} e^{in}} = \dfrac{1-e^{iN}}{1-e^i}$$
$$ |S_N | = \left | \dfrac{1-e^{iN}}{1-e^i} \right | = \dfrac{|1-e^{iN}|}{|1-e^i|}$$
Now, $|z+w|^2 \leq (|z| + |w|)^2 \forall z, w \in \mathbb{C}$, so:
$$|S_N| \leq \dfrac{(1+1)^2}{|1-e^i|} = \dfrac{4}{|1-e^i|}$$
So $|S_N|$ is bounded and we're done.
