<s>[INST] Infinite limit of trigonometric series The value of $\displaystyle\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))$ is
(A) $\sin^4x$
(B) $\sin^2x$
(C) $\cos^2x$
(D) does not exist
My attempt:
$$\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))=$$
$$=(\sin^4x+\frac{1}{4}16\sin^4x\cos^4 x+\cdots+\frac{1}{4^n}\sin^4(2^{n-1}x)\cos^4(2^{n-1}x)$$
i could not solve further.Any hint will be useful.
 [/INST] Hint: Note that $\sin^2 \theta - \sin^4\theta = \sin^2\theta(1-\sin^2\theta) = \sin^2\theta\cos^2\theta = \dfrac{1}{4}\sin^2 2\theta$. 
Hence, $\sin^4\theta = \sin^2\theta - \dfrac{1}{4}\sin^2 2\theta$. 
Applying that here gives us $\dfrac{1}{4^k}\sin^4(2^k x) = \dfrac{1}{4^k}\sin^2(2^k x) - \dfrac{1}{4^{k+1}}\sin^2(2^{k+1}x)$. 
So, we need to compute the limit of $\displaystyle\sum_{k = 0}^{n}\dfrac{1}{4^k}\sin^4(2^k x)$ $=\displaystyle\sum_{k = 0}^{n}\left[\dfrac{1}{4^k}\sin^2(2^k x) - \dfrac{1}{4^{k+1}}\sin^2(2^{k+1}x)\right]$.
This is a telescoping sum.
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