$\lim_{n\to\infty} \frac{1}{\log(n)}\sum _{k=1}^n \frac{\cos (\sin (2 \pi \log (k)))}{k}$ What tools would you gladly recommend me for computing precisely the limit below? Maybe a starting point?
$$\lim_{n\to\infty} \frac{1}{\log(n)}\sum _{k=1}^n \frac{\cos (\sin (2 \pi  \log (k)))}{k}$$
 A: 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.
A: I would start with this:
$$0\leq \left|\frac{1}{\log(n)}\sum_{k=0}^n\frac{\cos(\sin(2\pi\log(k)))}{k}\right|\leq \frac{1}{\log{n}}\sum_{k=0}^n\frac{1}{k}$$
A: Using the standard estimate for the harmonic numbers,
$$
\left| \frac{1}{\log n} \sum_{k=1}^n \frac{\cos(\sin(2\pi \log(k)))}{k}\right| \leq \frac{1}{\log n} \sum_{k=1}^n\frac{1}{k} = \frac{1}{\log n}(\log(n) + \gamma + O(1/n)).
$$
So in the limit,
$$
\left| \lim_{n\to \infty}\frac{1}{\log n} \sum_{k=1}^n \frac{\cos(\sin(2\pi \log(k)))}{k} \right| \leq 1.
$$
This is a pretty elementary bound, so you were probably already aware of it, but hopefully it helps.
