Evaluating $\lim_{n\to \infty}(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n})^{1/n}$ Find:
$$\lim_{n\to \infty}\left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n}$$
I was wondering,is there a formula for multiplication of sines,when angles are in arithmetic progression(likewise there is formula for summation of sines).I also tried this question by taking log of both sides and then applying L Hospital rule.But it is becoming messy.Can someone please assist me in solving this question? 
 A: We have
$$\lim_{n\to\infty}\ln \left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n} =\lim_{n\to\infty}\sum_{k=1}^{n-1} \frac{1}{n}\ln \sin\left(\frac{k}{n}\cdot \frac{\pi}{2} \right) =\hspace{1cm}\int_{0}^{1} \ln \sin\frac{\pi t}{2} dt   $$
Hence $$\lim_{n\to\infty} \left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n} =e^{\int_0^1\ln \sin\frac{\pi t}{2} dt}  $$
But it can be calculated that $$\int_0^1\ln \sin\frac{\pi t}{2} dt =-\ln2$$ therefore $$\lim_{n\to\infty} \left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n} =e^{-\ln 2 }=\frac{1}{2}  $$
A: First we have
\begin{align}
\lim_{n\to \infty}\frac1{n}\ln{\left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}\cdots\sin\frac{(n-1)\pi}{2n}\right)}
&=\lim_{n\to \infty}\frac1{n}\sum\limits_{k=0}^{n-1}\ln{\left(\sin\frac{k\pi}{2n}\right)}
\\
&=\frac{2}{\pi}\int_0^{\pi/2}\ln{\sin{x}}\:dx
\\
&=\frac{2}{\pi}\int_0^{\pi/2}\ln{\cos{x}}\:dx\tag{$y=\pi/2-x$}
\end{align} 
Let $I=\int_0^{\pi/2}\ln{\sin{x}}\:dx$. Then
\begin{align}
2I&=\int_0^{\pi/2}\ln{\sin{x}}\:dx+\int_0^{\pi/2}\ln{\cos{x}}\:dx
\\
&=\int_0^{\pi/2}\ln{\sin{x}\cos{x}}\:dx
\\
&=\int_0^{\pi/2}(\ln{\sin{2x}}-\ln{2})\:dx
\\
&=\int_0^{\pi/2}\ln{\sin{2x}}\:dx-\frac{\pi\ln{2}}{2}
\\
&=\frac1{2}\int_0^{\pi}\ln{\sin{x}}\:dx-\frac{\pi\ln{2}}{2}
\\
&=\frac1{2}\left(\int_0^{\pi/2}\ln{\sin{x}}\:dx+\int_{\pi/2}^{\pi}\ln{\sin{x}}\:dx\right)-\frac{\pi\ln{2}}{2}
\\
&=\frac1{2}\left(\int_0^{\pi/2}\ln{\sin{x}}\:dx+\int_{0}^{\pi/2}\ln{\cos{x}}\:dx\right)-\frac{\pi\ln{2}}{2}
\\
\end{align}
So  $I=-\dfrac{\pi\ln{2}}{2}$. And
\begin{align}
\lim_{n\to \infty}\left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}\cdots\sin\frac{(n-1)\pi}{2n}\right)^{1/n}&=\lim_{n\to \infty}e^{\frac1{n}\sum\limits_{k=0}^{n-1}\ln{\left(\sin\frac{k\pi}{2n}\right)}}
\\
&=e^{\frac{2}{\pi}I}=e^{-\ln{2}}
\\
&=\frac1{2}
\end{align}
A: $\bf{My\; Solution::}$ USing $\bf{n^{th}}$ root of Unity, $\displaystyle$ So Let $x= (1)^{\frac{1}{n}}\Rightarrow x^n - 1=0$
So $$x^n-1 = (x-1)\cdot (x-\alpha)\cdot (x-\alpha^2)\cdot.........(x-\alpha^{n-1})\;,$$
Where $$\displaystyle \alpha^r = \cos\left(\frac{2r\pi}{n}\right)+i\sin \left(\frac{2r\pi}{n}\right)\;,$$ and $r=0,1,2,3,.....(n-1)$
So $$\displaystyle \frac{x^n-1}{x-1} = (x-\alpha)\cdot (x-\alpha^2)\cdot.........(x-\alpha^{n-1})$$
Now Using The formula
$$\bullet\; x^n-1 = (x-1)\cdot \left(x^{n-1}+x^{n-2}+x^{n-3}+........+x^2+x+1\right)$$
So $$\displaystyle \left(x^{n-1}+x^{n-2}+x^{n-3}+........+x^2+x+1\right) = (x-\alpha)\cdot (x-\alpha^2)\cdot.........(x-\alpha^{n-1})$$
Now Put $x=1$ in above equation and taking Modulus on both side, We get
$$\displaystyle 1 = \left|(1-\alpha)\cdot (1-\alpha^2)\cdot.........(1-\alpha^{n-1})\right| = \left|(1-\alpha)\right|\cdot \left|(1-\alpha^2)\right|\cdot .........\left|(1-\alpha^{n-1})\right|$$
So we get
$$\displaystyle n= \left|1-\cos\left(\frac{2\pi}{n}\right)-i\sin \left(\frac{2\pi}{n}\right)\right|\cdot \left|1-\cos\left(\frac{4\pi}{n}\right)-i\sin \left(\frac{4\pi}{n}\right)\right|.........\left|1-\cos\left(\frac{2(n-1)\pi}{n}\right)-i\sin \left(\frac{2(n-1)\pi}{n}\right)\right|$$
So $$\displaystyle n = 2\sin\left(\frac{\pi}{n}\right)\cdot 2\sin\left(\frac{2\pi}{n} \right)....2\sin\left(\frac{(n-1)\pi}{n}\right)$$
So $$\displaystyle \sin\left(\frac{\pi}{n}\right)\cdot \sin\left(\frac{2\pi}{n} \right)....\sin\left(\frac{(n-1)\pi}{n}\right) = \frac{n}{2^{n-1}}$$
Now Replace $n\rightarrow 2n\;,$ We get 
$$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right) = \frac{2n}{2^{2n-1}}$$
So $$\displaystyle \lim_{n\rightarrow \infty}\left[\sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right)\right]^{\frac{1}{n}} = \lim_{n\rightarrow \infty}\left(\frac{2n}{2^{2n-1}}\right)^{\frac{1}{n}}$$
$$\displaystyle  = \frac{\lim_{n\rightarrow \infty}(2n)^{\frac{1}{n}}}{\lim_{n\rightarrow \infty}(2)^{2-\frac{1}{n}}} = \frac{1}{2^2}$$
For Calculation of $$\lim_{n\rightarrow \infty}(2n)^{\frac{1}{n}}$$
Let $$\displaystyle L=\lim_{n\rightarrow \infty}(2n)^{\frac{1}{n}}\Rightarrow \ln(L)=\lim_{n\rightarrow \infty}\frac{\ln(2n)}{n}$$
Now Using $$\bf{L,Hopital\; Rule}\;,$$ we get $$\ln(L)=0\Rightarrow L=e^{0}=1$$
I did not Understand Where I have Done Wrong, Plz anyone explain me, Thanks
