Convergence of $\sum_{k = 1}^{\infty} \cos (\log k)$ and $\sum_{k = 1}^{\infty} \sin (\log k)$ I was trying to determine if
$$\sum_{k = 1}^{\infty} \cos (\log k)$$
(with $k = 1, 2, \ldots$) converges. $\log$ is the natural logarithm.
I tried the ratio test,
$$\lim_{k \to \infty} \frac{\cos \left[ \log ( k + 1 ) \right]}{\cos \left[ \log ( k ) \right]} = 1$$
but it is not useful to determine the convergence.
According to Wolfram Alpha, it diverges. How to prove it?
And what about $\displaystyle \sum_{k = 1}^{\infty} \sin (\log k)$?
What I can empirically observe is that, regardless of how big $\log k$ can become, its cosine will always be a number $x$ such that $-1 \leq x \leq 1$. But I can't immediately see that this leads to an infinite sum.

Trivially, the term test is:
$$\lim_{k \to \infty} \cos (\log k)$$
and this limit is not defined, due to the cosine. Maybe this is sufficient to state that the series diverges.
 A: Suppose the series converges
$$
\sum_{k=1}^\infty \cos(\log k) = s \in \Bbb R
$$
and let $s_n = \sum_{k=1}^n \cos(\log k)$.
This would imply
$$
\cos(\log(n)) = s_{n} - s_{n-1} \to s-s = 0
$$
by arithmetic of limits, but $\lim_{n\to\infty} \cos(\log n)$ doesn't exist.
The argument that $\sum \sin(\log k)$ diverges should be similar.
A: $$\log \lfloor e^{2\pi k} \rfloor = \log( e^{2\pi k}+O(1))=
\log e^{2\pi k} +\log(1+ O(e^{-2\pi k}))=2\pi k+O(e^{-2\pi k})$$
Thus $$\lim_{k\to \infty}\cos(\log \lfloor e^{2\pi k} \rfloor )=1$$
Similarly $$\lim_{k\to \infty}\sin(\log \lfloor e^{2\pi (k+1/4)} \rfloor )=1$$
A: For $k$ sufficiently big, $\log(k+1)-\log k<\frac\pi2$. On the other hand, $\log k$ is unbounded. Therefore, $\log k$ "walks" to $\infty$ in steps so small that it will hit every interval $[2m\pi-\frac\pi4,2m\pi+\frac\pi4]$ with $m\gg0$. That is, we will have $\cos \log k\ge \frac{\sqrt 2}2$ infinitely often.
A: Suppose that $a_n =\cos (\log n )\to 0$ then of course $a_{10^k} =\cos k \to 0.$ But it is well known  that the last is not true.
