Divergence of $\sum_{n=1}^{\infty}\frac{a \cos(b \ln n)+b \sin(b \ln n)}{n^a}$

Question

Divergence of $$\sum_{n=1}^{\infty} \frac{a \cos(b \ln n) + b\sin(b \ln n)}{n^a}$$ where $0<a<1, b>0$ are constants.

My try: $$\cos(x)\geq 1-\frac{x^2}{2} , \forall x$$ $$\sin (x)\geq -x , \ \forall x$$ $$a\cos(b \ln n)+b\sin(b \ln n)\geq a\left(1-\frac{(b\ln \ n)^2}{2}\right)-b^2 \ln\ n$$

How to bound the given summation?

Answer 1

Hint: Let's try something simpler: $\sum_{n=1}^{\infty}\dfrac{\cos(\ln n)}{n^a}.$ For each positive integer $m,$ let $E_m$ be the set of integers $n$ such that

$$2\pi m <\ln n < \pi/3+2\pi m.$$

Then

$$\sum_{n\in E_m}\frac{\cos(\ln n)}{n^a} > |E_m|\cdot\frac{1}{2}\cdot \frac{1}{(e^{\pi/3+2\pi m})^a}.$$