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Find : $$\sqrt{ \frac{1}{2}}\sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}}\sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} +\frac{1}{2}\sqrt{\frac{1}{2}}}}} \\ \ldots \infty$$

Note by Kushagraa Aggarwal
4 years ago

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Let $$a_1 = \sqrt{\dfrac{1}{2}}$$ and let $$a_{n+1} = \sqrt{\dfrac{1}{2} + \dfrac{1}{2}a_n}$$ for integers $$n \ge 1$$. We want the value of $$\displaystyle\prod_{n = 1}^{\infty}a_n$$.

Note that $$a_1 = \cos \left(\dfrac{\pi}{4}\right)$$ and if $$a_n = \cos\left(\dfrac{\pi}{2^{n+1}}\right)$$ then $$a_{n+1} = \sqrt{\dfrac{1}{2} + \dfrac{1}{2}\cos\left(\dfrac{\pi}{2^{n+1}}\right)} = \cos\left(\dfrac{\pi}{2^{n+2}}\right)$$.

Therefore, by induction, we have $$a_n = \cos\left(\dfrac{\pi}{2^{n+1}}\right)$$ for all integers $$n \ge 1$$.

Let $$P_N = \displaystyle \prod_{n = 1}^{N} a_n = \displaystyle \prod_{n = 1}^{N}\cos\left(\dfrac{\pi}{2^{n+1}}\right)$$. Using the identity $$\sin \theta \cos \theta = \dfrac{1}{2}\sin 2\theta$$ repeatedly, we have:

$$P_N \sin\left(\dfrac{\pi}{2^{N+1}}\right) = \sin\left(\dfrac{\pi}{2^{N+1}}\right)\displaystyle \prod_{n = 1}^{N}\cos\left(\dfrac{\pi}{2^{n+1}}\right) = \dfrac{1}{2^N}\sin\left(\dfrac{\pi}{2}\right) = \dfrac{1}{2^N}$$.

Therefore, $$P_N = \dfrac{1}{2^N}\csc\left(\dfrac{\pi}{2^{N+1}}\right)$$ for all integers $$N \ge 1$$.

For $$x \approx 0$$ we have $$\csc x = \dfrac{1}{x} + O(x)$$. Hence, $$P_N = \dfrac{1}{2^N}\left[\dfrac{2^{N+1}}{\pi} + O\left(\dfrac{\pi}{2^{N+1}}\right)\right] = \dfrac{2}{\pi} + O(2^{-2N})$$.

As $$N \to \infty$$ we have $$2^{-2N} \to 0$$. Therefore, $$\displaystyle\prod_{n = 1}^{\infty}a_n = \lim_{N \to \infty}P_N = \dfrac{2}{\pi}$$. · 4 years ago

thanks a lot!. it was actually not easy to get to substitute $$\frac{1}{\sqrt{2}}$$ by $$cos(\frac{\pi}{4})$$ · 3 years, 12 months ago

If you don't mind, can you please share the source of this problem?

Thanks! · 3 years, 12 months ago

it was asked in a mock test. · 3 years, 12 months ago

Are you talking about an IIT-JEE mock test? If so, FIITJEE? · 3 years, 12 months ago

Nope , Vidyamandir Classes. · 3 years, 12 months ago