For $n \geq 1,$ let $G_n$ be the geometric mean of $\{\sin (\frac{\pi}{2} \frac{k}{n}) : 1 \leq k \leq n\}.$ Then $\lim_{n \rightarrow \infty} G_n?$

Question

For $n \geq 1,$ let $G_n$ be the geometric mean of $\{\sin (\frac{\pi}{2} \frac{k}{n}) : 1 \leq k \leq n\}.$ Then $\lim_{n \rightarrow \infty} G_n?$

How do I find out $G_n$? Please help.

Answer 1

Since we have $\log G_{n} = \frac{1}{n} \sum_{k=1} \log \sin \left( \frac{\pi k}{2n}\right)$, as $n$ ties to infinity, it converges to $$ \lim_{n\to \infty} \log G_{n} = \int_0^1 \log\sin\left(\frac{\pi x}{2}\right)dx$$ which can be done in various way. See here.