Analyzing convergence of series with sine and cosine Analyze the convergence of the following series:
$$\displaystyle\sum\frac{\cos{n}}{\sqrt{n}+\cos{n}}$$
$$\displaystyle\sum\frac{\sin{n}}{\sqrt{n}+\cos{n}}$$
I tried to use the direct comparison test to prove their absolute value diverges, for example:
$$\displaystyle|\frac{\cos{n}}{\sqrt{n}+\cos{n}}| \geq \frac{|\cos{n}|}{|\sqrt{n}|+|\cos{n}|} \geq \frac{\cos{n}}{\sqrt{n}+1}$$
But from there I can't get any additional information. Which test should I be using in this case?
 A: First we have the following asymptotic expansion
\begin{align}
  \frac{\cos n}{\sqrt{n}+\cos n} & = \frac{\cos n}{\sqrt{n}}\frac{1}{1+\frac{\cos n}{\sqrt{n}}}\\
   & =\frac{\cos n}{\sqrt{n}}\left( 1-\frac{\cos n}{\sqrt{n}}+\frac{\cos^2 n}{n}+O(\frac{1}{n^{3/2}})\right)\\
&=\frac{\cos n}{\sqrt{n}}+\frac{\cos (2n)}{2n}-\frac{1}{2n}+O(\frac{1}{n^{3/2}})
\end{align}
Now we are going to prove that the series $\sum \frac{\cos n}{\sqrt{n}}$ and $\sum \frac{\cos (2n)}{2n}$ are convergent using the Abel transform.  This will imply that the series $\sum \frac{\cos n}{\sqrt{n}+\cos n}$ is divergent.
We put $$S_n=\sum_{k=0}^n\cos k.$$
We have for every $n$
\begin{align*}
  |S_n| & = |\Re\left(\sum_{k=0}^n e^{ik}\right)|\\
   & =|\frac{e^{i(n+1)/2}}{e^{i/2}}\frac{\sin((n+1)/2)}{\sin(1/2)}|\\
&\leq\frac{1}{\sin(1/2)}.
\end{align*}
Then for every $n,p\geq1$ 
\begin{align*}
  |\sum_{k=n}^{n+p}\frac{\cos n}{\sqrt{n}}| & = |\sum_{k=n}^{n+p}\frac{S_k-S_{k-1}}{\sqrt{k}}|\\
   & =|\sum_{k=n}^{n+p}\frac{S_k}{\sqrt{k}}-\sum_{k=n-1}^{n+p-1}\frac{S_k}{\sqrt{k+1}}|\\
&=|\frac{S_{n+p}}{\sqrt{n+p}}-\frac{S_{n-1}}{\sqrt{n}}+\sum_{k=n}^{n+p-1}S_k\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)|\\
&\leq \frac{2}{\sin(1/2)\sqrt{n}}+\frac{1}{\sin(1/2)\sqrt{n}}
\end{align*}
Hence we deduce from the Cauchy convergence test that   the series $\sum \frac{\cos n}{\sqrt{n}}$  converges. 
We can show analogously that the series $\sum \frac{\cos (2n)}{2n}$ is convergent.
