Limit n tends to infinity

Question

How can i solve this: $$ \lim_{n\to\infty} \cos(1)\cos(0.5)\cos(0.25)\ldots \cos(1/2^n) $$

I tried using comlex numbers and logarithms but did'nt work out.Can anyone help please.

http://math.stackexchange.com/questions/455995/finding-the-limit-lim-limits-n-to-infty-cos-frac-x-2-cdot-cos-frac — lab bhattacharjee, Dec 26 '13 at 13:40

score 2 · Answer 1 · answered Dec 26 '13 at 13:20

2

hint: multiply $sin \dfrac{1}{2^n}$

answered Dec 26 '13 at 13:20

chenbai

7,581

Brilliant hint. Why this post was reported as low quality is beyond me... – Lost1 Dec 26 '13 at 13:50

score 1 · Answer 2 · answered Dec 26 '13 at 13:22

1

Note that $$ \sin(x) = 2 \sin(x/2) \cos(x/2) = 2^2\sin(x/4) \cos(x/2) \cos(x/4) = \ldots = 2^n \sin(x/2^n)\prod_{k=1}^n \cos(x/2^k) $$ Hence, $$ \lim_{n\to\infty} \prod_{k=1}^{n} \cos(x/2^k) = \lim_{n\to \infty} \frac{\sin(x)}{2^n \sin(x/2^n)} = \frac{\sin(x)}{x} \lim_{n\to\infty} \frac{x/2^n}{\sin(x/2^n)} $$ Now use the fact that $$ \lim_{z \to 0}\frac{\sin(z)}{z} = 1 $$ Now plug in $x=1$.

answered Dec 26 '13 at 13:22

Prahlad Vaidyanathan

31,554

1

so answer is sin1 ? – Devgeet Patel Dec 26 '13 at 13:28
very cool answer – Don Larynx Dec 26 '13 at 13:35

score 0 · Answer 3 · edited Apr 13 '17 at 12:20

0

We have the infinite product expansion

$$\frac{\sin x}{x}=\prod _{k=1}^\infty \cos\left(\frac{x}{2^k}\right)$$

This result was first found by Francois Viète in 1593. For a proof, see also this math.se question.

edited Apr 13 '17 at 12:20

Community

1

answered Dec 26 '13 at 13:33

Listing

13,937

Limit n tends to infinity

3 Answers3