$a_n=(1-\cos (\ln n)) \ln n$ has subsequence converging to 0 Original Question:

For $n=1, 2, ...$, define $a_n=(1- \cos (\ln n)) \ln n$. Show that $a_n$ has a subsequence converging to $0$, and also a subsquence converging to $+\infty$. Show also that $a_{n+1}-a_n \rightarrow0$, and deduce that any positive number $\alpha$, there is a subsequence of $a_n$ that converges to $\alpha$.


Source: Alan F. Beardon, "Limits - A New Approach to Real Analyis", 1997

I want to focus on the first subquestion only, i.e. $a_n$ has a subsequence converging to $0$.
I really have no clue how to attempt this question. Any hint? Thanks.
 A: For every $k\in \mathbb{N}$ there exists a natural number $n_k$ such that
$$e^{2\pi k}\le n_k< e^{2\pi k}+1=e^{2\pi k}(1+e^{-2\pi k}).$$
Then, making use of  $\log(1+x)\le x$ for $x\ge 0,$ gives
$$2\pi k \le \log n_k\le   2\pi k+\log (1+e^{-2\pi k})  \le 2\pi k +e^{-2\pi k}.$$
Hence $$\log n_k=2\pi k +\delta_k,\quad 0\le \delta_k\le e^{-2\pi k}.$$
Making use of $1-\cos x\le x$ for $x\ge 0,$  implies
$$1-\cos(\log n_k)=1-\cos \delta_k\le \delta_k\le e^{-2\pi k}
.$$
Next
$$0\le [1-\cos(\log n_k)]\log n_k\le e^{-2\pi k}\log n_k\le e^{-2\pi k}(2\pi k+1).$$
We get $$\lim_k [1-\cos(\log n_k)]\log n_k=0$$ by the squeeze theorem.
By considering the sequence $m_k\in \mathbb{N}$ such that
$$e^{(2k+1)\pi}\le m_k< e^{(2k+1)\pi}+1$$ we can prove that
$$\lim_k [1-\cos(\log m_k)]\log m_k=\infty.$$
The property $\displaystyle\lim_n (a_n-a_{n+1})=0$ follows from $$\lim_{x\to \infty}{d\over dx}[1-\cos(\log x )]\log x=0.$$ It can also be verified straightforward using trigonometry and $|\sin x|\le |x|.$
This property implies that the sequence $a_n$ is dense in $[0,\infty).$ Indeed, assume that an interval of $[a,a+\delta],$ for $a,\delta >0,$ does not contain any point $a_n.$ There is $n_0$ such that $$|a_n-a_{n+1}|<{\delta \over 2},\qquad n\ge n_0.$$ This implies that for $n\ge n_0$ all the points $a_n$ are less than  $a$ or all
$a_n$ are greater than $a+\delta.$ This gives a contradiction with the existence of two subsequences, one  tending to $0$, the other to $\infty.$
