$\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....\forall x \in \mathbb{R}$ is convergent?

Question

Is $\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....\forall x \in \mathbb{R}$ convergent? If it is convergent find the sum of the series.

Gives series $$\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....=\sum _{n=1}^\infty \frac{(-1)^{n+1}}{2^{n-1}}\sin^n x$$

$$\left|\sum _{n=1}^\infty \frac{(-1)^{n+1}}{2^{n-1}}\sin^n x\right|\le \sum _{n=1}^\infty \frac{1}{2^{n-1}}=2$$ so is it convergent?

Answer 1

Let $a:=\sin x$ then the series can be written as $$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\sin^nx}{2^{n-1}}=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{a^n}{2^{n-1}}=a\sum_{n=1}^{\infty}(-1)^{n-1}\Big(\frac{a}{2}\Big)^{n-1}=a\frac{1}{1-(-a/2)}=\frac{2a}{2+a}$$ by the geometric series whenever $|a|<2$ which is equivalent to $|\sin x|<2$ which is true for all $x\in \mathbb{R}$.

Answer 2

It converges to the limit $\frac{\sin x}{1+\frac{1}{2}\sin x}$ ;$a=\sin x$ and $r$(common ratio)$=-\frac{1}{2}\sin x.$ Since $|a|<1$.