I'm trying to answer this having been sent a photo of the questions from someone doing A level Edexcel Maths and I'm struggling. Any advice would be appreciated.
Evaluate and justify the validity of your answer (2 marks) $$\sum_{n=1}^\infty \left(\frac{1}{2}\sin x\right)^n $$
My attempts so far:
When $n = 0$
$$\left(\frac{1}{2}\sin x\right)^n = 1$$
and
$$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$$
so:
$$\sum_{n=1}^\infty \left(\frac{1}{2}\sin x\right)^n = \sum_{n=0}^\infty \left(\frac{1}{2}\sin x\right)^n - 1 = \frac{1}{1- \frac{1}{2}\sin x} -1 = \frac{2}{2-\sin x}-1 = \frac{\sin x}{2-\sin x}$$
As far as I can see it seems to converge to between $-\frac{1}{3}$ and 1 depending on $x$. Is that it? Am I missing anything?