Convergence of $\sum\frac{|\cos n|}{n\log n}$

Question

I wonder if the series $$\sum_{n=1}^\infty\frac{|\cos n|}{n\log n}$$ converges.

I tried to applying the condensation test, getting $$\sum\frac{2^n|\cos 2^n|}{2^n\log{2^n}}=\sum\frac{|\cos 2^n|}{n\log 2}$$ but I don't know how to show it converges?

Am I in the right way?

Be careful with the limits of summation. also have a look here: http://math.stackexchange.com/questions/574503/infinite-series-sum-n-2-infty-frac1n-log-n — KonKan, Feb 17 '16 at 03:55

score 1 · Accepted Answer · answered Feb 17 '16 at 06:40

Note that

$$|\cos n| \geqslant \cos^2 n = \frac1{2} + \frac1{2}\cos(2n).$$

Hence

$$\sum_{k=2}^n \frac{|\cos k|}{k \log k}\geqslant \frac1{2}\sum_{k=2}^n \frac{1}{k \log k}+\frac1{2}\sum_{k=2}^n \frac{\cos(2k)}{k \log k}.$$

The series on the LHS diverges because the first sum on the RHS diverges by the integral test and the second sum converges by Dirichlet's test.

score 0 · Answer 2 · answered Feb 17 '16 at 04:15

I believe that the sum diverges since the average value of $|\cos(n)|$ is around $\frac12$ and $\sum_{n=2}^{\infty} \frac1{n \log n}$ diverges.

A proof could probably be made rigorous by noting that, if $|\cos n|$ is small, then $n$ is close to $\pi(k+\frac12)$ for some integer $k$, so that $|\cos(n-1)|$ and $|\cos(n+1)|$ would be bounded away from zero so that the sum of the three terms for $n-1, n, n+1$ would be greater than some multiple (about, say, 2/3 or 1/2) of the corresponding sum for $\frac1{n \log n}$.

Convergence of $\sum\frac{|\cos n|}{n\log n}$

2 Answers2