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The value of $$\lim_{n\to\infty}\left(\sin\frac{\pi}{2n}\cdot \sin\frac{2\pi}{2n}\cdot \sin\frac{3\pi}{2n}\cdots \sin\frac{(n-1)\pi}{2n}\right)^{\frac{1}{n}}$$ is equal to?

Can anyone remind me of any formula for such series involving sine and cosine. I tried taking the limit of $(n-1)^{th}$ term it tends to sine π which is $0$. so the series indeed converges. Now from here how do I proceed to determine the value of the limit of this whole term. Please explain.

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    Find the logarithm of the expression, then consider Riemann sums. I'm not sure how one would evaluate the resulting integral, though. – 2017-01-23
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    Is it $2n$ in the last denominator? – 2017-01-23
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    maybe try to express this in complex variable see if it helps? – 2017-01-23
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    Can you please give a hint of Feynman's trick @Open Ball . I have no idea of it. – 2017-01-23
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    @shadowkh I am not sure if it can be cleverly used here, but the way I imagined it, it would be the same as integrating by parts (which I don't currently see as a fruitful thing) – 2017-01-23
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    Dear Readers please note I am just an undergraduate . so please make hints very simple as I won't be able to understand if I haven't come across. – 2017-01-23
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    I think last term in the denominator should be 2n ideally – 2017-01-23
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    You can find methods to compute the integral here: http://math.stackexchange.com/questions/37829/computing-the-integral-of-log-sin-x – 2017-01-23
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    Use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$ like http://math.stackexchange.com/questions/465075/find-lim-limits-n-to-infty-frac1n-sum-limits2n-r-1-fracr-sq and many others and then http://math.stackexchange.com/questions/189590/can-int-0-pi-2-ln-sinx-dx-be-evaluated-with-complex-method/189663#189663 – 2017-01-23

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lab bhattacharjee told you what to do (Riemann Integral):

$\displaystyle \lim\limits_{n\to\infty}\prod\limits_{r=1}^n (\sin\frac{\pi r}{2n})^{\frac{1}{n}}=\exp \int\limits_0^1 \ln\sin\frac{\pi x}{2} dx=\frac{1}{2}$

Note:

$\displaystyle -\ln(2\sin\frac{\pi x}{2})=-i\pi\frac{x-1}{2}+\sum\limits_{k=1}^\infty\frac{e^{-i\pi x k}}{k}\enspace$ for $\enspace 0

Integration from $0$ to $1$ gives a real number on the left side und an imaginary number on the right side and therefore on both sides $0$ which means $\displaystyle \int\limits_0^1 \ln\sin\frac{\pi x}{2} dx=\int\limits_0^1 (-\ln 2)dx=-\ln 2\enspace$.

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    Actually I have not studied Riemann integral till now . just for motivation can u please go ahead with a few more steps .coz all I know is to try using normal integration in your above expression which I've no idea to manipulate around with.thanks – 2017-01-23
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    And the answer to the question is given as 1/4. – 2017-01-23
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    @shadow kh : e.g. https://en.wikipedia.org/wiki/Riemann_integral ; $1^{st}$ step: Logarithm . $2^{nd}$ step: look at that what *lab bhattacharjee* wrote and set $f(x):=\ln \sin \frac{\pi x}{2}$ . $3^{rd}$ step: exponentiation . The integral you will find somewhere in the literature. --- $\frac{1}{4}$ ? I will check. – 2017-01-23
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    @shadow kh : I wrote you an answer as a note above; $\frac{1}{4}$ is false. – 2017-01-23
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    Its possibe.thanks – 2017-01-23
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    @shadow kh :You are welcome :-) – 2017-01-23