Does $\sum_{n=2}^\infty\frac{\cos\ln\ln n}{\ln n}$ converge?

Question

$$\sum_{n=2}^\infty\frac{\cos\ln\ln n}{\ln n}$$ My idea is $$-\frac1{\ln n}\le\frac{\cos\ln\ln n}{\ln n}\le\frac1{\ln n}$$ But I don't know if $\sum\frac1{\ln n}$ converges.

Answer 1

The first term in the Euler-Macluarin Summation Formula is

$$\int_1^N \frac{\cos(\log\log x)}{\log(x)}\,dx=\int_{-\infty}^{\log\log N} e^{e^x}\cos(x)\,dx$$

which diverges as $N\to \infty$. Therefore, the series of interest diverges.

Answer 2

Off the top of my head and from my phone:

Cos(log log n) Is essentially constant for longer and longer stretches and sum 1/log n diverges so the sum diverges.

I'm sure this could be made rigorous for any function thst grows slowly like log log.

Answer 3

As proved here that:

The series $\sum_{n=1}^\infty f(n)$ and integral $\int_1^\infty f(x) \, dx$ converge or diverge simultaneously if $f'$ is absolutely integrable over $[1,\infty).$

In this question, $f(x)=\frac{\cos(\log\log x)}{\log(x)},x\geq2$ and $$f'(x)=-\frac{\sin(\log\log x)+\cos(\log\log x)}{x\log^2x},$$ which imples $$|f'(x)|\leq\frac{2}{x\log^2x},\quad x\geq2.$$ Hence $$\int_2^\infty|f'(x)|\, dx=\int_2^\infty\frac{2}{x\log^2x}\, dx<\infty.$$

By the result above, we can get $$\sum_{n=2}^\infty\frac{\cos(\log\log n)}{\log n}$$ is divergent. Actually, we have $$\int_2^\infty\frac{\cos(\log\log x)}{\log(x)}\,dx =\int_{\log\log2}^\infty e^{e^u}\cos u\,du,$$ which is divergent by Cauchy Convergent Principle.