# $\lim \limits_{n \to \infty}$ $\prod_{r=1}^{n} \cos(\frac{x}{2^r})$

$\lim \limits_{n \to \infty}$ $\prod_{r=1}^{n} \cos(\frac{x}{2^r})$

How do I simplify this limit?

I tried multiplying dividing $\sin(\frac{x}{2^r})$ to use half angle formula but it doesnt give a telescopic which would have simplified it.

• I changed the question. I am sorry. I got confused. Its product. – user242605 May 22 '15 at 6:01

$$\prod_{r=1}^{\infty} \frac{\cos (x/2^r) \sin (x/2^r)}{\sin (x/2^r)} = \prod_{r = 1}^{\infty} \frac{\sin (x/2^{r-1})}{2\sin(x/2^r)} = \lim_{r \to \infty} \frac{\sin x}{2^r \sin (x/2^r)}$$

Can you take it from here? Recall that $\lim_{t \to 0} \frac{\sin \alpha t}{t} = \alpha$.

• I am sorry. There is no summation. Its product of the terms. – user242605 May 22 '15 at 5:51
• I am stupid. After using half-angle formula, it cancels out terms. (I dont know whether that is called telescopic or not). The last limit will give me sin(x)/x as the answer. Thank you so much. – user242605 May 22 '15 at 6:05
• @user242605 ~*No such thing as a stupid question*~, as it goes. – MCT May 22 '15 at 6:06
• Does the cancellation of terms make the series telescopic? And I was stupid :( – user242605 May 22 '15 at 6:10
• @user242605 Yes, this is a telescopic product. – MCT May 22 '15 at 6:12

Hint: Using the identity $\sin 2t = 2\sin t \cos t$, we have

$$\prod_{r = 1}^n \cos \frac{x}{2^r} = \prod_{r = 1}^n \frac{\sin \frac{x}{2^{r-1}}}{2\sin \frac{x}{2^r}} = \frac{\sin x}{2^n \sin \frac{x}{2^n}}$$

Now use the fact that $(\sin t)/t \to 1$ as $t\to 0$ to show that the limit of the above expression is $(\sin x)/x$.

• Thank you so much. I used the above identity but could not see that terms cancelled out. – user242605 May 22 '15 at 6:09

$$\lim \limits_{n \to \infty} \left(\prod_{r=1}^{n} \cos\left(\frac{x}{2^r}\right)\right)=\lim \limits_{n \to \infty} \left(\frac{\sin(x)\csc\left(\frac{x}{2^n}\right)}{2^n}\right)=\frac{\sin(x)}{x}$$