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 $\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ ?

Solution 1:

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....

Solution 2:

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, $$\int_{0}^{1}\log\sin x\,dx = -1-\sum_{k=1}^{+\infty}\sum_{j=1}^{+\infty}\frac{1}{j(2j+1)(k\pi)^{2j}},$$ $$\int_{0}^{1}\log\sin x\,dx = -1-\sum_{j=1}^{+\infty}\frac{\zeta(2j)}{j(2j+1)\pi^{2j}} = -\left(1+\frac{1}{18}+\frac{1}{900}+\frac{1}{19845}+\ldots\right).$$