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Limit of product of cos(rπ/2n) to power 1/n | JEE

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\lim_{n\to\infty}\left(\cos\dfrac{\pi}{2n}\,\cos\dfrac{2\pi}{2n}\,\cos\dfrac{3\pi}{2n}\cdots\cos\dfrac{(n-1)\pi}{2n}\right)^{1/n}

equals

1

2

0

\dfrac{1}{2}

correct

Solution

Step 1: Let

y

be the limit and take logarithms to convert the product into a sum:

\ln y=\lim_{n\to\infty}\dfrac{1}{n}\sum_{r=1}^{n-1}\ln\cos\dfrac{r\pi}{2n}.

Step 2: Recognise a Riemann sum. With

x=\dfrac{r}{n}

and spacing

\dfrac{1}{n}

, as

n\to\infty

this sum becomes a definite integral over

[0,1]

\ln y=\int_{0}^{1}\ln\cos\dfrac{\pi x}{2}\,dx.

Step 3: Substitute

\dfrac{\pi x}{2}=u

dx=\dfrac{2}{\pi}\,du

, with

u:0\to\dfrac{\pi}{2}

\ln y=\dfrac{2}{\pi}\int_{0}^{\pi/2}\ln\cos u\,du.

Step 4: Apply the standard result

\displaystyle\int_{0}^{\pi/2}\ln\cos u\,du=-\dfrac{\pi}{2}\ln 2

. Hence

\ln y=\dfrac{2}{\pi}\left(-\dfrac{\pi}{2}\ln 2\right)=-\ln 2\ \Rightarrow\ y=\dfrac12.

Correct answer: (4)

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