How to prove the series $\sum\limits_{n=1}^\infty\frac1n\sin(\ln n)$ diverges

As proved here, you can always apply the integral test to $\sum_{n \geqslant 1} f(n)$ even if $f$ is not monotone as long as

$$\tag{*}\int_1^\infty|f'(x)| \, dx < +\infty$$

In this case, $f'(x) = -\sin (\ln x)/x^2 + \cos(\ln x)/x^2$. Since $|f'(x)| \leqslant 2/x^2$, the condition (*) is met.

It then follows that the series diverges along with the integral $\int_1^\infty f(x) \, dx$.

(2018-11-25) "Amusing" revenge downvote, three months later.

Since it seems that everyone feels compelled to rush to the comparison with an integral, let us present a simpler, low-tech, proof (which, additionally, is the OP's idea).

Start with the elementary fact that $$\sin(\log n)\geqslant\frac12$$ for every $n$ such that $a_k\leqslant n\leqslant b_k$ for some $k$, with $$a_k=\lceil e^{2k\pi+\pi/6}\rceil\qquad b_k=\lfloor e^{2k\pi+5\pi/6}\rfloor$$ Thus, the slice of the series from $a_k$ to $b_k$ can be lower bounded as follows: $$\sum_{n=a_k}^{b_k}\frac1n\sin(\log n)\geqslant\frac12\sum_{n=a_k}^{b_k}\frac1n\geqslant\frac12(b_k-a_k+1)\frac1{b_k}$$ Since $a_k\to\infty$ and the RHS above converges to $$\frac12(1-e^{-2\pi/3})\ne0$$ the series of interest diverges.

Let $$a_n = \frac{\sin \ln n}{n}$$ For $n > 0$, let's take

$$b_n = a_n - \int_n^{n+1}\frac{\sin\ln x}{x} dx = \int_0^1 \left(a_n - \frac{\sin \ln(n+t)}{n+t}\right) dt $$ Since for each $\alpha \in ]0,1]$, we can apply MVT to find a $c \in [0,\alpha]$

$$\left|a_n - \frac{\sin\ln(n+\alpha)}{n+\alpha}\right| = \left|\frac{\cos\ln(n+c) - \sin\ln(n+c)}{(n+c)^2}\right|\alpha \le \frac{\sqrt{2}\alpha}{n^2}$$ As a consequence, $$|b_n| \le \frac{\sqrt{2}}{n^2}\int_0^1 \alpha d\alpha = \frac{1}{\sqrt{2}n^2}$$ By the $p-$test, $\sum \frac{1}{\sqrt{2}n^2}$ is a convergent series. So by the comparison test, $\sum b_n$ is an absolutely converging series. Let $A = \sum b_n < \infty$

$$A= \lim_{N\to\infty}\left[\sum_{n=1}^N a_n - \sin\ln(N+1)\right] $$ Since the limit of $ \sin\ln(N+1)$ oscillates, then the limit of $\sum_{n=1}^N a_n$ has to oscillate to maintain a finite $A$. Hence $\sum_{n=1}^{\infty} a_n$ diverges.

NOTE: Answer was inspired by @achille hui