Evaluating $\lim\limits_{n \to \infty }\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}} $ Help me please with the limit:
$$\lim_{n \to \infty }\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\frac{\left | \sin2 \right |}{2}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}}$$
Thanks!
 A: $$\lim_{n\to\infty}\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\frac{\left | \sin2 \right |}{2}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}}  \leq \lim_{n\to\infty}{\dfrac1{\sqrt[n]{n!}}}=\lim_{n\to\infty}{\dfrac n{\sqrt[n]{n!}}}\cdot\frac{1}{n}=\lim_{n\to\infty} \frac{e}{n}=0.$$
Q.E.D. (we can safely apply d'Alembert criterion for  $\lim_{n\to\infty}{\dfrac n{\sqrt[n]{n!}}})$
A: $$\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\frac{\left | \sin2 \right |}{2}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}}  \leq \sqrt[n]{\dfrac1{n!}}$$ Now our good old Stirling's formula will do the job.
A: We have the following theorem, which is plain if you learned preliminary real analysis.

Theorem. If $a_n > 0$ and $\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \ell$, then $\displaystyle \lim_{n\to\infty} \sqrt[n]{a_n} = \ell$.
Proof. We only consider the case $\ell > 0$. The case $\ell = 0$ easily follows by slightly modifying the proof of the case $\ell > 0$.
Now for any $\epsilon \in (0, \ell)$, there exists $N = N(\epsilon)$ such that whenever $n \geq N$, we have
$$ \ell - \epsilon < \frac{a_{n+1}}{a_n} < \ell + \epsilon.$$
Multiplying this inequality for $n$ replaced by $N, N+1, \cdots, n-1$ for $n > N$, we have
$$ (\ell - \epsilon)^{n-N} \leq \frac{a_n}{a_N} < (\ell + \epsilon)^{n-N}.$$
Taking $n$-th root and letting $n\to\infty$, we obtain
$$ \ell - \epsilon \leq \liminf_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \sqrt[n]{a_n} \leq \ell + \epsilon.$$
Since $\epsilon > 0$ is arbitrary, both limsup and liminf coincide with common value $\ell$, hence follows the claim.

Now you can apply this directly to conclude that the limit is zero.
A: With AM-GM the sequence is smaller than $\frac{H_n}{n}$ with Hn is équivalent ln (n) , then the limite is zéro.
