Lockdown Math - Limits

$\begin{aligned} \lim_{n \to \infty} G_n & = \lim_{n \to \infty} \sqrt[n]{\prod_{k=1}^n \sin \frac {k\pi}{2n}} \\ & = \lim_{n \to \infty} \exp \left(\ln \left(\sqrt[n]{\prod_{k=1}^n \sin \frac {k\pi}{2n}}\right) \right) & \small \blue{\text{where }\exp (x) = e^x} \\ & = \exp \left(\lim_{n \to \infty} \frac 1n \sum_{k=1}^n \ln \left(\sin \frac {k\pi}{2n}\right)\right) & \small \blue{\text{By Riemann sums}} \\ & = \exp \left(\blue{\int_0^1 \ln \left(\sin \frac {\pi x}2 \right) dx} \right) & \small \blue{\text{See note below.}} \\ & = \exp \left(\blue{-\ln(2)}\right) = \frac 12 = \boxed{0.5} \end{aligned}$

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Note:

$\begin{aligned} I & = \int_0^1 \ln \left(\sin \frac {\pi x}2\right) dx & \small \blue{\text{Let }\theta = \frac {\pi x}2 \implies d\theta = \frac \pi 2 dx} \\ & = \frac 2\pi \int_0^\frac \pi 2 \ln \left(\blue{\sin \theta} \right) d\theta & \small \blue{\text{By Euler's formula}} \\ & = \frac 2\pi \int_0^\frac \pi 2 \ln \left(\blue{\frac {e^{i\theta}-e^{-i\theta}}{2i}} \right) d\theta \\ & = \frac 2\pi \int_0^\frac \pi 2 \ln \left(e^{i\theta} - e^{-i\theta} \right) d\theta - \frac 2 \pi \int_0^\frac \pi 2 \ln (2i)\ d\theta \\ & = \frac 2\pi \int_0^\frac \pi 2 \ln \left(\frac {1 -e^{-2i\theta}}{e^{-i\theta}} \right) d\theta - \frac 2 \pi \ln (2i) \left( \frac \pi 2 - 0 \right) \\ & = \blue{\frac 2\pi \int_0^\frac \pi 2 \ln \left(1 -e^{-2i\theta}\right) d\theta} - \frac 2 \pi \int_0^\frac \pi 2 \ln \left(e^{-i\theta} \right) d\theta - \ln (2i) & \small \blue{\text{Let }\phi = 2\theta \implies d\phi = 2 d\theta} \\ & = \blue{\frac 1\pi \int_0^\pi \ln \left(1 -e^{-i\phi}\right) d\phi} - \frac 2 \pi \int_0^\frac \pi 2 \left(-i\theta \right) d\theta - \ln (2i) & \small \blue{\text{By Maclaurin series}} \\ & = \blue{\frac 1\pi \int_0^\pi -\left(\sum_{n=1}^\infty \frac {e^{-n i\phi}}n \right) d\phi} + \frac {i\theta^2} \pi \bigg|_0^\frac \pi 2 - \ln 2 - \ln \left(e^{\frac {i\pi} 2}\right) \\ & = \frac 1\pi \left[\sum_{n=1}^\infty \frac {e^{-n i\phi}}{n^2i} \right]_0^\pi + \frac {i\pi}4 - \ln 2 - \frac {i\pi} 2 \\ & = - \frac i\pi \sum_{n=1}^\infty \frac {(-1)^n-1}{n^2} - \frac {i\pi}4 - \ln 2 \\ & = \frac {2i}\pi \sum_{n=0}^\infty \frac 1{(2n+1)^2} - \frac {i\pi}4 - \ln 2 \\ & = \frac {2i}\pi \left(\blue{\zeta (2)} - \frac \blue{\zeta (2)}4 \right) - \frac {i\pi}4 - \ln 2 & \small \blue{\text{where }\zeta(\cdot) \text{ is the Riemann zeta function.}} \\ & = \frac {2i}\pi \left(\frac 34 \times \blue{\frac {\pi^2}6}\right) - \frac {i\pi}4 - \ln 2 & \small \blue{\text{and }\zeta(2) = \frac {\pi^2}6} \\ & = \frac {i\pi}4 - \frac {i\pi}4 - \ln 2 \\ & = - \ln 2 \end{aligned}$

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References:

• Riemann sums
• Euler's formula
• Riemann zeta function

This is not an algebra problem, but a calculus problem . The value of the limit is

$e^{\displaystyle \int_0^1 \ln \sin \frac{πx}{2}dx}=\dfrac{1}{2}=\boxed {0.5}$ .