Desmos seems to say it diverges, oscillating between $1.5$ and $1.7$, but I can't seem to prove it. Here's the question.
Prove if $\sum_{k=2}^\infty\frac{\sin(k)}{\log(k)}$ converges or diverges.
Here's my attempt so far. Ratio test and root test gives the limit going to $1$, so they are inconclusive. Cauchy Condensation Test doesn't apply here as well.
I've tried to show that the sequence of partial sums is not Cauchy. That is, letting $s_n = \sum_{k=2}^n\frac{\sin(k)}{\log(k)}$, $s_{N+n} = \sum_{k=2}^n\frac{\sin(k)}{\log(k)}$, one has:
\begin{align}
|s_{N+n} – s_n| &=\left| \frac{\sin(N+n)}{\log(N+n)} + \cdots +\frac{\sin(N+1)}{\log(N+1)} \right| \\
& \geq \frac{1}{N+n}|\sin(N+n) + \cdots +\sin(N+1)|
\end{align}
But after this I struggle to find a hard bound on the above inequality that is not dependent on $n$.
Comparing this series against another series doesn't work as well. $\sum \frac{1}{\log(k)}$ diverges and $\sum \frac{\sin(k)}{k}$ converges to $\frac{\pi-1}{2}$, but $\sum \frac{\sin(k)}{k}$ is smaller and $\sum \frac{1}{\log(k)}$ is bigger, which says nothing about the series in question.
Hints?
Edit: The integral test is not in my course, and from what I know about it, you need your function to be positive.
Best Answer
Notice that \begin{align}\left\lvert \sum_{k=2}^N \sin k\right\rvert&=\left\lvert\sum_{k=2}^N \frac{e^{ik}-e^{-ik}}{2i}\right\rvert=\left\lvert \frac{e^{(k+1)i}-e^{2i}}{2i(e^i-1)}-\frac{e^{-(k+1)i}-e^{-2i}}{2i(e^{-i}-1)}\right\rvert=\\&=\left\lvert\frac{(\sin(k+1)-\sin 2)(\cos 1-1)-(\cos(k+1)-\cos 2)\sin 1}{(\cos 1-1)^2+\sin^21}\right\rvert\le\\&\le \frac{5}{(\cos 1-1)^2+\sin^21}\end{align}
Therefore, by Dirichlet's test, the series $\sum_{k=2}^\infty\frac{\sin k}{\log k}$ converges.