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.
 A: 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.
A: HINT:
$$\sin^2y-\sin^4y=\sin^2y(1-\sin^2y)=\dfrac{\sin^22y}4$$
$y=x\implies$  $$\sin^2x-\sin^4x=\dfrac{\sin^22x}4$$
$y=2x\implies$  $$\dfrac{\sin^22x-\sin^42x}{4^1}=\dfrac{\sin^22x}{4^2}$$
Set $y=4x,2^rx$ and add to get
$$\sin^2x-\sum_{r=0}^n\dfrac{\sin^4(2^rx)}{4^r}=\dfrac{\sin^4(2^{n+1}x)}{4^{n+1}}$$  
$$\lim_{n\to\infty}\dfrac{\sin^4(2^{n+1}x)}{4^{n+1}}=0$$
A: $$\begin{align} \sin^2 (x) - \sin^4(x) &= \sin^2 (x) \cos^2(x) = \frac14 \sin^2(2x), \\
 \frac14 \sin^2 (2x) - \frac14 \sin^4(2x) &= \sin^2 (2x) \cos^2(2x) = \frac{1}{4^2} \sin^2(4x),\\
&\cdots \\
\frac{1}{4^{n}} \sin^2 (2^{n}x) - \frac{1}{4^{n}} \sin^4(2^{n}x) &= \sin^2 (2^{n}x) \cos^2 (2^{n}x)= \frac{1}{4^{n+1}} \sin^2(2^{n+1} x). 
\end{align}
$$
So , the limit 
$$ \begin{align}
\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx)) &= \lim_{n\to\infty}\left( \sin^2 (x) - \frac{1}{4^{n+1}} \sin^2(2^{n+1} x) \right) \\
&=  \sin^2 (x) .
\end{align}$$
If you do not know this. For your multiple choice question:


*

*Set $x = 0$. the limit (if exists) is $0$, so C is false.

*For all $n$, $|\frac{1}{4^n}\sin(2^n x)| \leqslant \frac{1}{4^n}$ , thus by dominated convergence theorem, the sequence uniformly converges to a continues function, say $g(x)$. D is false.

*Choose an $x_0 \in (0,\frac{\pi}{2})$ such that $\frac13 <\sin^4(2x_0) < 1$, then $\frac14 \sin^4(2x_0)$ is strictly superior to the sum of the absolute value of the rest of the terms (which exists since the series is absolutely convergent and is inferior to $\frac{1}{12}$ by a simple comparison). So $|g(x_0) - \sin^4(x_0)| \geqslant \frac14 \sin^4(2x_0) - \sum|\mbox{rest of the terms}| > 0$.   So A is false.


So B is the only correct choice.
A: Simple solution without any calculation
Firstly, the limit exists because the absolute value of the terms are bounded by a geometric progression, so the partial sums form a Cauchy sequence and hence converges. Secondly, substituting $x = \frac{π}{2}$ and $x = \frac{π}{4}$ eliminates the wrong answers immediately.
