2
$\begingroup$

I have been told that $\lim_{m\to \infty}\left(\cos (x/2) + \cos (x/2^2) + \cdots + \cos (x/2^m)\right)=\frac{x}{\sin (x)}$

I tried working with the power series of $\sin(x)$ and $\cos(x)$ but it becomes messy pretty soon. I was hoping that there might be some trick to arrive at such a conclusion. I would be grateful for any suggestions.

  • 4
    Sorry to break the spell but the limit of the sum in the LHS is +infinity, for every x.2012-11-25

3 Answers 3

3

This answers what I think is probably the question the OP has in mind, which is not at present the question they asked.

Consider, for every $x$ and every $n\geqslant0$, $ f_n(x)=\prod_{k=1}^n\cos(x/2^k). $ Then, using repeatedly the identity $\sin(2t)=2\sin(t)\cos(t)$, a recursion on $n\geqslant0$ yields $ 2^n\sin(x/2^n)f_n(x)=\sin(x). $ When $n\to\infty$, for every $x\ne0$, $2^n\sin(x/2^n)\to x$. Furthermore, $f_n(0)=1$ for every $n$ hence, for every $x$, $ \lim_{n\to\infty}f_n(x)=\frac{\sin x}x. $

  • 0
    thank you :) i guess you must be right2012-11-25
0

You can use $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$, then you should be able to see that the sum on the left side has a remarkable similarity to the fact that $(1+x)(1+x^2)(1+x^4)(1+x^8)...=1+x+x^2+x^3+x^4+...=\frac{1}{1-x}$

  • 0
    Sorry but I am unable *to see that the sum on the left side has a remarkable similarity to the fact* you mention2012-11-25
0

Try to substitute $\cos(x/2^n)=\sin(x/2^{n-1})/(2\sin(x/2^n))$, then you obtain a telescopic product.