# Find the limit when $x = \pi / 6$

Let $a_0 = 1$ and $a_n = a_{n-1}(\cos \frac{x}{2^n})$ .Find the limit $\{a_n\}$ when $x = \pi / 6$ . (i.e. find the $\lim_{n \to \infty} a_n$)

My try : I solved it but the solution was time-consuming .

$\frac{a_n}{a_{n-1}} = \cos \frac{x}{2^n} \to \frac{a_n}{a_0} = \frac{a_n}{a_{n-1}} \times \frac{a_{n-1}}{a_{n-2}} \times \dots \times\frac{a_1}{a_0} = \cos {\frac{x}{2^n}} \times \cos {\frac{x}{2^{n-1}}} \times \dots \times \cos {\frac{x}{2}} \to (2^n \sin{\frac{x}{2^n}})(a_n) = \sin x \to a_n = \frac{\sin x}{2^n\sin{\frac{x}{2^n}}} = \frac{\frac{x}{2^n}}{\sin{\frac{x}{2^n}}} \times \frac{\sin x }{x} \to \lim_{n \to \infty } a_n = 1 \times \frac{\sin{\frac{\pi}{6}}}{\frac{\pi}{6}} = \frac{3}{\pi}$

Note : I did some calculations by hand and didn't write them .

• That's actually the most direct solution. I'd be surprised if you found an easier one.
– dxiv
Aug 7 '17 at 19:20
• @dxiv You're right but it was only 1 minute time for solving this . Aug 7 '17 at 19:22
• That's a bit tight, indeed, yet it depends on the context and prior work. Once you "see" the telescoping (or have seen it before, maybe) the rest is straightforward.
– dxiv
Aug 7 '17 at 19:31
• @dxiv I agree with you . The telescoping is the main idea . Aug 7 '17 at 19:49

You already know that $$a_n = \cos \left({\frac{x}{2^n}}\right) \times \cos \left({\frac{x}{2^{n-1}}}\right) \times \dots \times \cos \left({\frac{x}{2}}\right) = \prod_{i=1}^n \cos \frac{\theta}{2^i}$$
Therefore $\displaystyle \lim_{n\to\infty} a_n = \prod_{i=1}^{\infty} \cos \dfrac{\theta}{2^i}$, if the product exists.
$$\prod_{n=1}^\infty \cos \frac{\theta}{2^n} = \frac{\sin \theta}{\theta}$$
Using this, we immediately get that the answer is $\dfrac{\sin (\pi/6)}{\pi/6} = \dfrac3{\pi}$.