<s>[INST] Find $\lim_{n\to\infty}\sin^4x+\frac{1}{4}\sin^42x+.....+\frac{1}{4^n}\sin^42^nx$ $\lim_{n\to\infty}\sin^4x+\frac{1}{4}\sin^42x+.....+\frac{1}{4^n}\sin^42^nx$
I feel that this question will be solved by telescoping series but i cannot express it as telescoping series.
$\sin^4x=(\frac{1-\cos2x}{2})^2=\frac{1+\cos^22x-2\cos2x}{4}$
I am not able to solve it further.
The options are $(A)\sin^4x,(B)\sin^2x,(C)\cos^2x$(D)does not exist
 [/INST] Given $$\sin^4 x = \sin^2 x(1-\cos^2 x) = \sin^2 x-\frac{1}{4}\sin^2(2x)$$
So $$\sin^4(2x) = \sin^2(2x)-\frac{1}{4}\sin^2(4x)$$
So $$\sin^4(4x) = \sin^2(4x)-\frac{1}{4}\sin^2(8x)$$
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So $$\sin^4(2^n x) = \sin^2(2^n x)-\frac{1}{4}\sin^2(2^{n+1}x)$$
$$\lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{1}{2^k}\sin^4(2^k x) = \lim_{n\rightarrow \infty}\bigg[\sin^2 x-\frac{1}{4^{n+1}}\sin^2 (2^{n+1})\bigg]=\sin^2 x.$$
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