KAIST POW 2012-7 : Product of sines

Product of sines

Let $X$ be the set of all positive real number $c$ such that $$\frac{\prod_{k=1}^{n-1}{\sin\frac{k\pi}{2n}}}{c^n}$$ converges as $n$ goes to infinity. Find the infimum of $X$.

Proof. Since $\sin\frac{\pi}{2}=1$,  we can say $\frac{\prod_{k=1}^{n-1}{\sin\frac{k\pi}{2n}}}{c^n}=\frac{\prod_{k=1}^{n}{\sin\frac{k\pi}{2n}}}{c^n}$

Define $B_n =\frac{\prod_{k=1}^{n-1}{\sin\frac{k\pi}{2n}}}{c^n}$. And taking logarithm to the sequence, then we define

$C_n = \log\biggl[\frac{\prod_{k=1}^{n-1}{\sin\frac{k\pi}{2n}}}{c^n}\biggr] = \sum_{k=1}^n \log(\sin\frac{k\pi}{2n})-n\log c =\biggl[\sum_{k=1}^n \log(\sin\frac{k\pi}{2n}) \frac{1}{n} -\log c\biggr]n$.

By definition of Riemann sum, $\lim_{n\to \infty} \sum_{k=1}^n \log (\sin\frac{k\pi}{2n})\frac{1}{n} = \int_0^1 \log(\sin\frac{k\pi}{2n})dx =\log (\frac{1}{2})$.

So, $\lim_{n\to \infty} \biggl[\sum_{k=1}^n \log(\sin \frac{k\pi}{2n})\frac{1}{n}-\log c \biggr] = \log \frac{1}{2c}$.

Since $\lim_{n\to \infty} n = \infty$,

if $c > \frac{1}{2}$, then $\log \frac{1}{2c} <0$ and  $\lim_{n\to \infty} C_n = -\infty$.

That is,  $\lim_{n\to \infty} B_n=\lim_{n\to \infty} e^{C_n} = 0$.

if $c < \frac{1}{2}$, then $\log \frac{1}{2c} >0$ and  $\lim_{n\to \infty} C_n = \infty$.

That is,  $\lim_{n\to \infty} B_n = \lim_{n\to \infty} e^{C_n} = \infty$.

Therefore, the infimum of $X$ is $\frac{1}{2}$