lim-x-sin-pi-2n-sin-2pi-2n-sin-3pi-2n-sin-n-1-pi-n-1-n-

Question Number 163658 by mathlove last updated on 09/Jan/22

\lim_{x \to \infty} \sqrt[n]{\sin \frac{π}{2 n} \times \sin \frac{2 π}{2 n} \times \sin \frac{3 π}{2 n} \dots . \times \sin \frac{(n - 1) π}{n}} = ?

Answered by Ar Brandon last updated on 09/Jan/22

A = \lim_{n \to \infty} \sqrt[n]{\sin \frac{π}{2 n} \times \sin \frac{2 π}{2 n} \times \sin \frac{3 π}{2 n} \dots . \times \sin \frac{(n - 1) π}{n}}

\ln A = \lim_{n \to \infty} \frac{1}{n} \ln (\underset{k = 1}{\prod^{n - 1}} \sin (\frac{k π}{2 n})) = \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n - 1}} \ln (\sin (\frac{k π}{2 n}))

= \int_{0}^{1} \ln (\sin (\frac{π}{2} x)) d x = \frac{2}{π} \int_{0}^{\frac{π}{2}} \ln (\sin u) d u

= \frac{2}{π} (- \frac{π \ln 2}{2}) = - \ln 2 \Rightarrow A = e^{\ln (\frac{1}{2})} = \frac{1}{2}

Answered by qaz last updated on 09/Jan/22

\lim_{n \to \infty} \sqrt[n]{\sin \frac{π}{2 n} \cdot \sin \frac{2 π}{2 n} \cdot \dots \cdot \sin \frac{(n - 1) π}{n}}

= \exp \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{2 n - 2}} lnsin \frac{k π}{2 n}

= e^{\int_{0}^{2} lnsin \frac{π}{2} xdx}

= \frac{1}{4}

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