Proving $\sum_{n=1}^\infty \cos(2n)\tan\left(\frac{1}{n}\right)$ doesn't absolutely converge

Question

I am trying to test whether the following series absolutely converges/converges/diverges: $$\sum_{n=1}^\infty \cos(2n)\tan\left(\frac{1}{n}\right)$$

I was able to prove it converges using Dirichlet's test.

I am trying to prove it does not converge absolutely, hence: $$\sum_{n=1}^\infty |\cos(2n)\tan\left(\frac{1}{n}\right)|$$

Diverges. I know $|\cos(2n)\tan\left(\frac{1}{n}\right)|\leq \tan\left(\frac{1}{n}\right)$ for every $n$, and I was able to prove $\sum_{n=1}^\infty \tan\left(\frac{1}{n}\right)$ diverges by using the comparison test to $\sum_{n=1}^\infty 1/n$, but cannot understand how to go forward. (as I'm looking for a divergent series that is smaller equal rather than bigger equal)

Thank you!

Answer 1

Since $\frac{2 \pi}{3} > 2$, every interval of the form $\left( 2k \pi - \frac{\pi}{3}, 2k \pi + \frac{\pi}{3} \right)$, where $k \geqslant 1$, contains an integer of the form $2n$. In this situation we have that

$$\cos(2n) \tan \frac{1}{n} \geqslant \frac{1}{2} \cdot \frac{1}{n} = \frac{1}{2n} \geqslant \frac{1}{2k \pi + \frac{\pi}{3}}$$

Summing this over $k = 1, 2, 3, \ldots$ we get that our sum is greater than $\displaystyle \sum_{k=1}^{\infty} \frac{1}{2k \pi + \frac{\pi}{3}}$, which is obviously divergent.

Answer 2

We have $|\cos(2n)\ge \cos^2(2n).$ Moreover $$\cos(4n)=2\cos^2(2n)-1$$ As $$\sum \cos(4n)\tan(1/(2n))$$ is convergent and $$\sum \tan(1/(2n))=\infty $$ we get $$\sum \cos^2(2n)\tan(1/(2n)=\infty$$ Therefore $$\sum |\cos(2n)|\tan(1/(n)\ge \sum |\cos(2n)|\tan(1/(2n))=\infty$$

Remark The same method can be applied to $$\sum |\cos(an)|\tan(b/n)=\infty ,\qquad a,b>0$$