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
$$ u_{n}\geq n! \prod_{k=1}^n \frac{2x}{\pi k}$$
$$ u_{n} \geq \prod_{k=1}^n \frac{2x}{\pi}$$
If $x=\pi/2$, $u_{n}\geq1$, $\sum u_{n}$ is divergent.
What about the case $x\in[1,\pi/2[$ ?