# try this one!

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
6 years, 6 months 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, $\displaystyle\prod_{n = 1}^{\infty}a_n = \lim_{N \to \infty}P_N = \dfrac{2}{\pi}$.

- 6 years, 6 months ago

thanks a lot!. it was actually not easy to get to substitute $\frac{1}{\sqrt{2}}$ by $cos(\frac{\pi}{4})$

- 6 years, 6 months ago

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

Thanks!

- 6 years, 6 months ago

it was asked in a mock test.

- 6 years, 6 months ago

Are you talking about an IIT-JEE mock test? If so, FIITJEE?

- 6 years, 6 months ago

Nope , Vidyamandir Classes.

- 6 years, 6 months ago

Oh yeah, 2/pi, Viete's formula.

- 6 years, 6 months ago