Find the limit of $(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ Could you tell me how to find $\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ ?
few thoughts on the first one:

$$\ln \left((\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}} \right)=\frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right)$$

This is just a Riemann sum, and thus its limit is $$\int_{0}^1 \ln(\sin(x)) dx$$

This is an improper integral though, so the RS approach might not be best, but I think it is a convergent improper integral, since $\int_0^1 \ln(x)dx $ is convergent..

Maybe someone can take over....
As $x\rightarrow 0$ , $\sin(x) \rightarrow x$ 

So $ \sin(\frac{1}{n}) \rightarrow \frac{1}{n}$ as $n \rightarrow \infty$ 

The product $ [ \sin(\frac{1}{n}) \cdot \sin(\frac{2}{n}) \cdot ... \cdot \sin(\frac{n}{n}) ] \rightarrow \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdot...\frac{n}{n} =
\frac{n!}{n^n}$ 

As $n! = \sqrt{2 \cdot \pi \cdot n} \cdot (\frac{n}{e})^n  $ 

So $\frac{n!}{n^n} = \sqrt{2\pi n} \cdot (\frac{1}{e})^n $

$\lim_{ n \rightarrow \infty}$ of $[ \sin(\frac{1}{n}) \cdot \sin(\frac{2}{n}) \cdot ... \cdot \sin(\frac{n}{n}) ] ^ \frac{1}{n} \rightarrow \frac{1}{e}$

Since $\sin(\frac{m}{n})$ is always positive  if $0&lt; m \le n$ and less than $\frac{m}{n}$ .
 The limit will be less than (1/e)