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 $\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ ?
 A: Yes, some care is needed in order to prove the convergence of the Riemann sum to the corresponding integral, from which
$$\lim_{n\to +\infty}\left(\prod_{k=1}^{n}\sin\frac{k}{n}\right)^{1/n}=\exp\left(\int_{0}^{1}\log\sin x dx\right)$$
immediately follows. The integral is just a bit less than $-1$, and we can use the Weierstrass product of the sine function to write the integral as a fast-converging series. Since:
$$ \sin x = x\prod_{k=1}^{+\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)$$
holds uniformly for $x\in[0,1]$, we have:
$$\int_{0}^{1}\log\sin x\,dx = -1+\sum_{k=1}^{+\infty}\left(-2-2k\pi\operatorname{arctanh}\frac{1}{k\pi}+\log\left(1-\frac{1}{\pi^2 k^2}\right)\right),$$
or, using the Taylor series of the logarithm and the hyperbolic arctangent,
$$\int_{0}^{1}\log\sin x\,dx = -1-\sum_{k=1}^{+\infty}\sum_{j=1}^{+\infty}\frac{1}{j(2j+1)(k\pi)^{2j}},$$
$$\int_{0}^{1}\log\sin x\,dx = -1-\sum_{j=1}^{+\infty}\frac{\zeta(2j)}{j(2j+1)\pi^{2j}} = -\left(1+\frac{1}{18}+\frac{1}{900}+\frac{1}{19845}+\ldots\right).$$
A: 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....
A: 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 $0< m \le n$ and less than $\frac{m}{n}$ .
 The limit will be less than (1/e)
