$$y=\lim_{n\to \infty}\left(\sin π/2n \times \sin 2π/2n \times... \sin(n-1)π/2n\right)$$ Find y. I took log on both the sides and I got $$\ln y= \int_0^1 {\ln \sin πx/2\, dx}$$ but I'm stuck here.
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1Maybe you can find some interesting answers here:http://math.stackexchange.com/questions/690644/evaluate-int-0-pi-2-log-cosxdx – Olivier Oloa Aug 19 '15 at 22:00
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1Yeah, that was helpful, Thanks. – exilednick Aug 19 '15 at 22:15
2 Answers
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If we wish to evaluate the limit of interest by use of Riemann sums, then we can proceed as follows. Let $y_n$ be the product defined by
$$y_n=\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{2n}\right) \tag 1$$
Taking the logarithm of both sides of $(1)$ reveals
$$\log y_n=\sum_{k=1}^{n}\log\left(\sin\left(\frac{k\pi}{2n}\right)\right) \tag 2$$
where we used the fact that $\sin \pi/2=1$, thereby permitting our extending the limit on the sum to $n$. Note that we can write the Riemann sum of the integral
$$\begin{align} \int_0^{\pi/2}\log\left(\sin x\right)\,dx&=\lim_{n\to \infty}\,\frac{\pi/2}{n}\,\sum_{k=1}^{n}\log\left(\sin\left(\frac{k\pi}{2n}\right)\right)\\\\ &=-\frac12 \pi \log 2 \tag 3 \end{align}$$
Comparing the $(2)$ and $(3)$ we see that
$$\lim_{n\to \infty}\log y_n =-\infty$$
which implies therefore that
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}y_n=0}$$
Mark Viola
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Obviously to compute the closed form of the product is a chance, but there are at least two good reasons for which that limit is zero that require very few computations:
- $f(\theta)=\log\sin\theta$ is an improperly Riemann-integrable function over $\left(0,\frac{\pi}{2}\right)$;
- for any $\theta\in\left(0,\frac{\pi}{2}\right)$, we have $0<\sin\theta<\theta.$
Jack D'Aurizio
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