As N.S. shows in his answer, the ratio test quickly resolves the problem for all $x\neq 1$, so for now fix $x=1.$ Then $\displaystyle u_n = \prod_{k=1}^n k\sin(1/k).$
We show that $\displaystyle \lim_{n\to\infty} u_n = L> 0 $ so $\sum u_n$ diverges.
Since $\sin xconvergence criterion for infinite products, let us work with $v_n = 1/u_n$ instead. Let us remember that "convergence" of an infinite product demands the partial products tends to a finite non-zero limit. We compute $$v_n=\prod_{k=1}^n \frac{1}{k\sin(1/k)} = \prod_{k=1}^n (1+a_n)$$
where $\displaystyle a_n = \frac{1}{k\sin(1/k)}-1 >0.$ We know that $\displaystyle \prod_{k=1}^n v_n$ converges if and only if $\displaystyle \sum_{k=1}^n a_n$ converges. Using Taylor series for the $\sin$ term and expanding as geometric series we have $$a_n= \frac{1}{1-1/(6k^2) +\mathcal{O}(k^{-3})}-1= \frac{1}{6k^2} + \mathcal{O}\left(\frac{1}{k^3}\right)$$
so we find that $\sum a_n$ converges, so $\sum u_n$ diverges.