$\lim_{n\to\infty} \frac{1}{\log(n)}\sum _{k=1}^n \frac{\cos (\sin (2 \pi \log (k)))}{k}$

Let $f(x) = \frac{\cos(\sin(2\pi x))}{x}$, we have

$$f'(x) = - \frac{2\pi\cos(2\pi x)\sin(\sin(2\pi x)) + \cos(\sin(2\pi x))}{x^2} \implies |f'(x)| \le \frac{2\pi + 1}{x^2} $$ By MVT, for any $x \in (k,k+1]$, we can find a $\xi \in (0,1)$ such that

$$|f(x) - f(k)| = |f'(k + \xi(x-k))|(x-k) \le \frac{2\pi+1}{k^2}$$

This implies

$$\left| \int_k^{k+1} f(x) dx - f(k)\right| \le \frac{2\pi+1}{k^2}$$

As a result,

$$|\sum_{k=1}^n f(k) - \int_1^n f(x) dx| \le |f(n)| + \sum_{k=1}^{n-1} \left|f(k) - \int_k^{k+1} f(x)dx\right|\\ \le 1 + (2\pi + 1)\sum_{k=1}^{\infty} \frac{1}{k^2} = 1 + \frac{(2\pi + 1)\pi^2}{6} < \infty$$

As a result,

$$\lim_{n\to\infty}\frac{1}{\log n}\sum_{k=1}^{n}f(k) = \lim_{n\to\infty}\frac{1}{\log n}\int_1^n f(x) dx = \lim_{L\to\infty}\frac{1}{L}\int_0^L f(e^t) de^t\\ = \lim_{L\to\infty}\frac{1}{L}\int_0^L \cos(\sin(2\pi t)) dt $$ Since the integrand is periodic with period $1$, the limit at the right exists and equal to

$$\begin{align} & \int_0^1\cos(\sin(2\pi t))dt = \frac{1}{2\pi}\int_0^{2\pi}\cos(\sin(t)) dt = J_0(1)\\ \approx & 0.765197686557966551449717526102663220909274289755325 \end{align} $$ where $J_0(x)$ is the Bessel function of the first kind.

I would start with this:

$$0\leq \left|\frac{1}{\log(n)}\sum_{k=0}^n\frac{\cos(\sin(2\pi\log(k)))}{k}\right|\leq \frac{1}{\log{n}}\sum_{k=0}^n\frac{1}{k}$$