Calculus II series I'm trying to find convergence or divergence of $$\sum_{n=1}^\infty\frac{\cos(\log n)}{n}\text{.}$$I've tried to use the squeeze theorem, however, my professor said that $\log(n)$ grows too slowly for the numerator to be assumed to be zero. I don't see a solution with nth term, limit comparison or direct comparison. Any suggestions would be great.

The series does not converge because it oscillates too much. Let $$S_m = \sum_{n=1}^m \frac{\cos \ln n}{n}$$ be the sequence of partial sums. Then for arbitrary $N > 0$, we can find $m_1, m_2$ with $|S_{m_1} - S_{m_2}| > 1$, so the partial sums continue oscillating by a constant amount no matter how far out you go.
(In real analysis terms, $(S_m)$ does not converge because it is not Cauchy.)
To prove this, given any $N>0$, choose $k$ such that $e^{(k-\frac13)\pi} > N$ and consider the sum
$$\sum_{n = \exp((k-\frac13)\pi}^{\exp((k+\frac13)\pi)} \frac{\cos \ln n}{n}.$$
On this interval, we have $(k - \frac13)\pi \le \ln n \le (k + \frac13)\pi$ and therefore either $\cos \ln n \ge \frac12$ or $\cos \ln n < -\frac12$ according to whether $k$ is even or odd. The two cases are symmetric, so let's assume $k$ is even.
In that case, each term of the sum is at least $\frac{1}{2n}$, and we have $$\sum_{n=a}^b \frac{1}{2n} \approx \frac12 \left(\ln b - \ln a\right)$$ with an error that goes to $0$ as we choose large enough $a, b$. (We know this from an asymptotic estimate of the harmonic numbers (Wikipedia) or from a comparison to the integral of $\frac{1}{2x}$ between $a$ and $b$.)
In this case, $a = \exp((k-\frac13)\pi)$ and $b = \exp((k+\frac13)\pi)$, so $\frac12(\ln b - \ln a) = \frac\pi3$. If $k$ is chosen so that our estimates are precise enough (and we can choose $k$ to be as large as we want) then
$$\sum_{n = \exp((k-\frac13)\pi}^{\exp((k+\frac13)\pi)} \frac{\cos \ln n}{n} > \frac\pi3 - 0.001 > 1.$$
