TITLE: Does this tricky series converge?
QUESTION [13 upvotes]: $$\sum_2 \frac{\cos(\log{n})}{n\log{n}}$$
The naive attempt is to use dirichlet's test to falsely claim that $\cos(\log{n})$ has bounded partial sums, but I don't think it works.
I am also trying a difference of sum and integral type of strategy but am not sure where to go from it.  
Finally, as A.S pointed out below in the comments, the integral test does not apply, since cos(logn) does not behave monotonically... 
Any ideas are welcome.
Thanks,

REPLY [9 votes]: First examine the corresponding integral:
$$\int_2^{\infty} \frac{\cos(\log(x))}{x \log(x)}dx = \int_{log(2)}^{\infty} \frac{\cos(u)}{u} du = \left[ \frac{\sin(u)}{u} \right]_{log(2)}^{\infty} + \int_{log(2)}^{\infty} \frac{\sin(u)}{u^2}du $$ this is convergent.
Next we look at the derivative of the function 
$$\begin{align}f(x) &= \frac{\cos(\log(x))}{x \log(x)} \\
f'(x) &= \frac{-1/x \cdot\sin(\log(x))\cdot x \log(x) + \cos(\log(x))(\log(x)+1)}{(x \log(x))^2}\\
 |f'(x)| &< \frac{2}{x^2}\end{align}$$ so we can compare the terms of sum and the integral over intervals of length 1:
$$\begin{align} \left| \frac{\cos(\log(k))}{k \log(k)} - \int_k^{k+1} \frac{\cos(\log(x))}{x \log(x)} dx \right| &\leq \int_k^{k+1} | f(k) - f(x) | dx \\ &\leq \max_{k<x<k+1} | f'(x) | < \frac{2}{k^2} \end{align}$$
Since the integral and the sum of the differences both are convergent, the original series is convergent.