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
 A: Hint :
$$\sin^4x =\dfrac18 \cos 4x-\frac12 \cos 2x+\frac38$$
A: Using the hint given by Almot1960 we have $\sin^4 x =\frac{3}{8} \color{red}{-\frac{1}{2} \cos(2x)} \color{blue}{ +\frac{1}{8} \cos (4x) }$ ...
\begin{eqnarray*}
\sin^4 x +\frac{1}{4} \sin^4 2x + \frac{1}{4^2} \sin^4 4x +\frac{1}{4^3} \sin^4 8x + \cdots \\
=\frac{3}{8} \color{red}{-\frac{1}{2} \cos(2x)} \color{blue}{ +\frac{1}{8} \cos (4x) } +\frac{3}{8} \frac{1}{4} \color{blue}{-\frac{1}{8} \cos(4x)} \color{red}{ +\frac{1}{32} \cos (8x) } +\frac{3}{8}\frac{1}{4^2} \color{red}{-\frac{1}{32} \cos(8x)} \color{blue}{ +\frac{1}{128} \cos (16x) } \cdots 
\end{eqnarray*}
So you were absolutely right , it is a telescoping sum & lots of the terms cancel. The terms in black are simply a geometric series
\begin{eqnarray*}
 \frac{3}{8} \left( 1+ \frac{1}{4}+\frac{1}{4^2}+ \cdots \right)= \frac{1}{2}
\end{eqnarray*}
So the sum is $ \frac{1- \cos(2x)}{2} =\sin^2 (x)$ and the answer is $\color{red}{(B)}$.
A: Consider : $\displaystyle \frac{\sin^{4}(2^{i}x)}{4^i} + \frac{\sin^{4}(2^{i+1}x)}{4^{i+1}} = \frac{1}{4^{i}}\big(\frac{1}{8}\cos(4\cdot2^{i}x) -\frac{1}{2}\cos(2\cdot2^{i}) + \frac{3}{8} + \frac{1}{32}\cos(2\cdot2^{i+1})-\frac{1}{8}\cos(2\cdot2^{i+1}) + \frac{3}{8}\big)$
Now you could subtract two terms in neighbour parts.
A: 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 $$\sin^4(2^n x) = \sin^2(2^n x)-\frac{1}{4}\sin^2(2^{n+1}x)$$
$$\lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{1}{2^k}\sin^4(2^k x) = \lim_{n\rightarrow \infty}\bigg[\sin^2 x-\frac{1}{4^{n+1}}\sin^2 (2^{n+1})\bigg]=\sin^2 x.$$
