It is unknown whether
$$
\sum_{n=1}^\infty\frac{1}{n^3\sin^2n}
$$
converges or not. The difficulty here is that convergence depends on the term $n\sin n$ not being too small, which in turn depends on how well $\pi$ can be approximated by rational numbers. It is possible that, if $\pi$ can be approximated `too well' by rationals, then this will diverge. See this MathOverflow question for a discussion of this particular series.

Another even simpler example of a sequence (no summation) for which it is not known whether it converges or not is
$$
x_n=\frac{1}{n^2\sin n}.
$$
We would expect this to tend to zero, but the proof is beyond what is currently known. Suppose that there were only finitely many rational numbers $p/q$ with $\vert p/q-\pi\vert\le q^{-3+\epsilon}$ (for any $\epsilon > 0$), then $x_n$ would tend to zero at rate $O(n^{-\epsilon})$. If, on the other hand, there were infinitely many rationals satisfying $\vert p/q-\pi\vert\le q^{-3-\epsilon}$, then infinitely many $x_n$ would be of order at least $n^\epsilon$, so it diverges. This can be expressed in terms of the irrationality measure of $\pi$. The sequence $x_n$ converges to zero if the irrationality measure of $\pi$ is less than 3, and diverges if it is geater than 3. Currently, the best known bound for the irrationality measure is that it is no more than about $7.6063$^{*} (see the link to the mathworld page above). It is expected that the irrationality measure of $\pi$ is 2 (it is known that all but a zero-measure set of real numbers have irrationality measure 2). Therefore, it is expected that $x_n$ tends to zero, but there is currently no proof of this.

[** The best known bound for the irrationality measure, as listed on the mathworld page, has been improved! It is now 7.10320533, according to the (not yet peer-reviewed) paper by Zeilberger and Zudlin, 2019. This is still much too small an improvement to say whether or not the sequence above converges.*]