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 $x\in[1,\pi/2[$ ?

$$\frac{u_{n+1}}{u_n}=\frac{(n+1)!\prod_{k=1}^{n+1} \sin\left(\frac{x}{k}\right)}{n!\prod_{k=1}^n \sin\left(\frac{x}{k}\right)}=(n+1)\sin(\frac{x}{n+1})$$
Thus, $\left|\lim_n \frac{u_{n+1}}{u_n}\right|=|x|$.
Now, ratio test solves all cases excepting $x=1$.
If $x=1$, I think the product is known, but can't remember a reference.
