ardeinduie

2022-01-23

What tools would you gladly recommend me for computing precisely the limit below? Maybe a starting point?
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\mathrm{log}\left(n\right)}\sum _{k=1}^{n}\frac{\mathrm{cos}\left(\mathrm{sin}\left(2\pi \mathrm{log}\left(k\right)\right)\right)}{k}$

trovabile4p

Let $f\left(x\right)=\frac{\mathrm{cos}\left(\mathrm{sin}\left(2\pi x\right)\right)}{x}$, we have
${f}^{\prime }\left(x\right)=-\frac{2\pi \mathrm{cos}\left(2\pi x\right)\mathrm{sin}\left(\mathrm{sin}\left(2\pi x\right)\right)+\mathrm{cos}\left(\mathrm{sin}\left(2\pi x\right)\right)}{{x}^{2}}⇒|{f}^{\prime }\left(x\right)|\le \frac{2\pi +1}{{x}^{2}}$
By MVT, for any $x\in \left(k,k+1\right]$ we can find a $\zeta \in \left(0,1\right)$ such that
$|f\left(x\right)-f\left(k\right)|=|{f}^{\prime }\left(k+\zeta \left(x-k\right)\right)|\left(x-k\right)\le \frac{2\pi +1}{{k}^{2}}$
This implies
$|{\int }_{k}^{k+1}f\left(x\right)dx-f\left(k\right)|\le \frac{2\pi +1}{{k}^{2}}$
As a result,
$|\sum _{k=1}^{n}f\left(k\right)-{\int }_{1}^{n}f\left(x\right)dx|\le |f\left(n\right)|+\sum _{k=1}^{n-1}|f\left(k\right)-{\int }_{k}^{k+1}f\left(x\right)dx|\le 1+\left(2\pi +1\right)\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{k}^{2}}$
$=1+\frac{\left(2\pi +1\right){\pi }^{2}}{6}<\mathrm{\infty }$
As a result,
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\mathrm{log}n}\sum _{k=1}^{n}f\left(k\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\mathrm{log}n}{\int }_{1}^{n}f\left(x\right)dx$
$=\underset{L\to \mathrm{\infty }}{lim}\frac{1}{L}{\int }_{0}^{L}\mathrm{cos}\left(\mathrm{sin}\left(2\pi t\right)\right)dt$

Damian Roberts

$0\le |\frac{1}{\mathrm{log}\left(n\right)}\sum _{k=0}^{n}\frac{\mathrm{cos}\left(\mathrm{sin}\left(2\pi \mathrm{log}\left(k\right)\right)\right)}{k}|\le \frac{1}{\mathrm{log}n}\sum _{k=0}^{n}\frac{1}{k}$

RizerMix

Using the standard estimate for the harmonic numbers, $|\frac{1}{\mathrm{log}n}\sum _{k=1}^{n}\frac{\mathrm{cos}\left(\mathrm{sin}\left(2\pi \mathrm{log}\left(k\right)\right)\right)}{k}|\le \frac{1}{\mathrm{log}n}\sum _{k=1}^{n}\frac{1}{k}$ $=\frac{1}{\mathrm{log}n}\left(\mathrm{log}\left(n\right)+\gamma +O\left(1/n\right)\right)$ So in the limit, $|\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\mathrm{log}n}\sum _{k=1}^{n}\frac{\mathrm{cos}\left(\mathrm{sin}\left(2\pi \mathrm{log}\left(k\right)\right)\right)}{k}|\le 1$ This is a pretty elementary bound, so you were probably already aware of it, but hopefully it helps.