Nature of the series $ \sum u_{n}, u_{n}=n!\prod_{k=1}^n \sin\left(\frac{x}{k}\right) $ Is the series $$ \sum u_{n}$$
$$ u_{n}=n!\prod_{k=1}^n \sin\left(\frac{x}{k}\right)$$ 
$$ x\in]0,\pi/2] $$
convergent or divergent?
We have:
$$ u_{n}\leq n!\prod_{k=1}^n \frac{x}{k}$$
$$ u_{n}\leq x^n$$
If $0<x<1$ the series is convergent.
$$ u_{n}\geq n! \prod_{k=1}^n \frac{2x}{\pi k}$$
$$ u_{n} \geq \prod_{k=1}^n \frac{2x}{\pi}$$
If $x=\pi/2$, $u_{n}\geq1$, $\sum u_{n}$ is divergent.
What about the case $x\in[1,\pi/2[$ ?

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 $\sin x<x$ we see $u_n<1.$ In order to apply a convergence 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. 
