<s>[INST] Prove two identities relating to series Show that:
$$(1)\sum_{n=1}^\infty\ln\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$$
$$(2). \sum_{n=1}^\infty\frac1{2^n}\tan \frac{x}{2^n}=\frac1x-\cot x$$
Thank you in advance.
NOTE:
The origial problem (1) I stated is $\sum_{n=1}^\infty\lim_{x\to \infty}\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$, which is a mistake as noted by the comments below.
 [/INST] For (1) Consider the product $P_N = \displaystyle\prod_{n = 1}^{N}\cos\dfrac{x}{2^n}$. Multiply by $\sin\dfrac{x}{2^N}$ and use the identity $\cos\theta\sin\theta = \dfrac{1}{2}\sin 2\theta$ repeatedly to get $P_N\sin\dfrac{x}{2^N} = \dfrac{1}{2^N}\sin x$. 
Therefore, $P_N = \dfrac{\sin x}{2^N\sin\frac{x}{2^N}} \to \dfrac{\sin x}{x}$ as $N \to \infty$. Hence, $\displaystyle\prod_{n = 1}^{\infty}\cos\dfrac{x}{2^n} = \dfrac{\sin x}{x}$. 
Take the natural log of both sides to get $\displaystyle\sum_{n = 1}^{\infty}\ln\left(\cos\dfrac{x}{2^n}\right) = \ln\left(\dfrac{\sin x}{x}\right)$, as desired. 
For (2) differentiate both sides of the result from (1) and multiply by $-1$. (Thanks David H for this suggestion). 
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