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.

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.
