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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.

Thanks in advance.

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  • $\begingroup$ How can you say that it is periodic.... $\endgroup$ – Jasser Sep 28 '14 at 16:37
  • $\begingroup$ well, in fact the sequence is not periodic, what is periodic is the cosine function, i made a mistake $\endgroup$ – Mike Sep 28 '14 at 16:45
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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.

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  • $\begingroup$ 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. $\endgroup$ – Jakobian Aug 1 '19 at 17:57
  • $\begingroup$ @Jakobian: the first part is needed to prove that the only possible limit, if existing, is one. $\endgroup$ – Jack D'Aurizio Aug 1 '19 at 17:59
  • $\begingroup$ Yes, consider this as an alternative approach. $\endgroup$ – Jakobian Aug 1 '19 at 20:00

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