TITLE: Find the limit of $(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$
QUESTION [13 upvotes]: Could you tell me how to find $\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ ?

REPLY [5 votes]: 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,
$$\int_{0}^{1}\log\sin x\,dx = -1-\sum_{k=1}^{+\infty}\sum_{j=1}^{+\infty}\frac{1}{j(2j+1)(k\pi)^{2j}},$$
$$\int_{0}^{1}\log\sin x\,dx = -1-\sum_{j=1}^{+\infty}\frac{\zeta(2j)}{j(2j+1)\pi^{2j}} = -\left(1+\frac{1}{18}+\frac{1}{900}+\frac{1}{19845}+\ldots\right).$$