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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 $n$?)

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No. $$ |\cos n|\ge\cos^2n=\frac{1+\cos(2\,n)}{2}. $$ $\displaystyle\sum_{n=1}^\infty\frac{\cos(2\,n)}{n}$ converges by Dirichlet's test, while $\displaystyle\sum_{n=1}^\infty\frac{1}{n}$ diverges.

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This shows convergence.Was the Q re-written? On Mar. 25 it asks about absolute convergence. – user254665 Mar 25 at 20:04
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@user254665 The series is the sum of a convergent series and a divergent series, so it is divergent. If it is divergent, it can't be absolutely convergent. – f'' Mar 25 at 20:35

$\cos(2n-1)\pi/2 =0$ 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.

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