The value of $$\lim_{n\to\infty}\left(\sin\frac{\pi}{2n}\cdot \sin\frac{2\pi}{2n}\cdot \sin\frac{3\pi}{2n}\cdots \sin\frac{(n-1)\pi}{2n}\right)^{\frac{1}{n}}$$ is equal to?
Can anyone remind me of any formula for such series involving sine and cosine. I tried taking the limit of $(n-1)^{th}$ term it tends to sine π which is $0$. so the series indeed converges. Now from here how do I proceed to determine the value of the limit of this whole term. Please explain.
lab bhattacharjee told you what to do (Riemann Integral):
$\displaystyle \lim\limits_{n\to\infty}\prod\limits_{r=1}^n (\sin\frac{\pi r}{2n})^{\frac{1}{n}}=\exp \int\limits_0^1 \ln\sin\frac{\pi x}{2} dx=\frac{1}{2}$
Note:
$\displaystyle -\ln(2\sin\frac{\pi x}{2})=-i\pi\frac{x-1}{2}+\sum\limits_{k=1}^\infty\frac{e^{-i\pi x k}}{k}\enspace$ for $\enspace 0<x<2$
Integration from $0$ to $1$ gives a real number on the left side und an imaginary number on the right side and therefore on both sides $0$ which means $\displaystyle \int\limits_0^1 \ln\sin\frac{\pi x}{2} dx=\int\limits_0^1 (-\ln 2)dx=-\ln 2\enspace$.