# Analyse of a simple trigonometric sequence

I can't figure out how i prove that the sequence $$W_{n} = \cos ({\ln({n}) )}$$ converge or diverge. I can say that the sequence is limited because $\mid W_{n}\mid = \mid \cos ({\ln({n}) )}\mid \hspace{0.1cm} \leq 1$, but at the same time the sequence is periodic and i don't know an idea to show that the sequence is monotonic, in the case that it converges, or why it diverges.

The limit does not exist. By assuming $$\lim_{n\to +\infty}\cos(\log n)=L$$ and replacing $$n$$ with $$n^2$$ we get that $$L$$ is a root of $$2z^2-1 = z,$$ but replacing $$n$$ with $$n^3$$ we get that $$L$$ is a root of $$4z^3-3z = z,$$ hence the only possibility is $$L=1$$. However, by taking $$n$$ as the closest integer to $$e^{(2k+1)\pi}$$ for $$k\in\mathbb{N}$$ (for example, integers between $$e^{(2k+1)\pi}$$ and $$e^{(2k+1.5)\pi}$$), we have that $$\{W_n\}_{n\in\mathbb{N}}$$ frequently belongs to a neighbourhood of $$-1$$ separated from $$1$$, hence the limit does not exist.
• The first part of the proof isn't really needed if you expand on the second one. If we show that $W_n$ belong infinitely many times to two separated neighbourhoods, one of $1$ and one of $-1$, then the claim easily follows. – Jakobian Aug 1 '19 at 17:57