TITLE: Is this series absolutely convergent (doesn't look like an easy problem)?
QUESTION [7 upvotes]: Is the series 
$$
\sum_{n=1}^{\infty} \frac{\cos n}{n}
$$
absolutely convergent?
(I've got a feeling that most probably it isn't due to the fact that for given 
$\varepsilon>0$ we can find infinitely many $n$ such that 
$|\cos n|>1-\varepsilon$, the problem is - how dense is the set of these $n$?)

REPLY [2 votes]: $\cos(2n-1)\pi/2 =0$ for all $n\ge1$
Set intervals as $[(2n-1)\pi/2-\pi/12, (2n-1)\pi/2+\pi/12]$ for each $n$. 
Then |cos$x$|<=|cos$5\pi/12$| for all $x$ in $[(2n-1)\pi/2-\pi/12, (2n-1) \pi/2+\pi/12]$. Otherwise $|\cos x|>|\cos5\pi/12|$.
Since the length of $[(2n-1) \pi/2-\pi/12, (2n-1)\pi/2+\pi/12]$ is smaller than $1$, at least one of two consecutive natural number is not in $[(2n-1)\pi/2-\pi/12, (2n-1)\pi/2+\pi/12]$, of which the value of $|\cos|$ is larger than $|\cos5\pi/12|$. 
Consequently the given series is larger than the infinite summation of $|\cos\pi/12|/2n$ for all $n\ge1$ so the given series diverges.
Or you can try other method using $\cos^2 x+\sin^2 x=1$ and proving the infinite summation of $cos^{2}n/n$ diverges. Other trigonometric formulas are also useful as well.