Question

The value of $\underset{n\to \infty }{lim}{(\tan \left(\frac{\pi }{2n}\right)\cdot \tan \left(\frac{2\pi }{2n}\right)\cdots \tan \left(\frac{(n-1)\pi }{2n}\right))}^{1/n}$ is

Answer 1

Let $P=\underset{n\to \infty }{lim}{(\tan \left(\frac{\pi }{2n}\right)\cdot \tan \left(\frac{2\pi }{2n}\right)\cdots \tan \left(\frac{(n-1)\pi }{2n}\right))}^{1/n}$
Taking $\ln$ both sides, we get
$\ln P=\underset{n\to \infty }{lim}\frac{1}{n}[\ln \tan \left(\frac{\pi }{2n}\right)+\ln \tan \left(\frac{2\pi }{2n}\right)+\cdots +\ln \tan \left(\frac{(n-1)\pi }{2n}\right)]\Rightarrow \ln P=\underset{n\to \infty }{lim}\sum _{r=1}^{n-1}\frac{1}{n}\ln \tan \left(\frac{\pi r}{2n}\right)=\underset{0}{\overset{1}{\int }}\ln \tan \left(\frac{\pi x}{2}\right)dx$
Putting $\frac{\pi x}{2}=y\Rightarrow dx=\frac{2}{\pi }dy,$ we have
$\ln P=\frac{2}{\pi }\underset{0}{\overset{\pi /2}{\int }}\ln \tan ydy\cdots \left(i\right)$

Now, let $I=\underset{0}{\overset{\pi /2}{\int }}\ln \tan ydy$
$=\underset{0}{\overset{\pi /2}{\int }}\ln \tan (\frac{\pi }{2}-y)dy(\because \underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx)$
$=\underset{0}{\overset{\pi /2}{\int }}\ln \cot ydy$
$=-\underset{0}{\overset{\pi /2}{\int }}\ln \tan ydy=-I$
So, $I+I=0\Rightarrow I=0$
From equation $\left(i\right)$,
$\ln P=0\Rightarrow P=e^0=1$