Answer
Converges to $\dfrac{3+\sin x}{8 + 2\sin x}$ for all $x$
Work Step by Step
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}$
The given series shows a convergent geometric series with first term, $a=\dfrac{1}{2}$ and common ratio $r =\dfrac{-1}{ 3+\sin x}$
$S=\dfrac{\dfrac{1}{2}}{1-(\dfrac{-1}{ 3+\sin x})}=\dfrac{3+\sin x}{2(4+\sin x)}=\dfrac{3+\sin x}{8 + 2\sin x}$
Hence, the given series converges to $\dfrac{3+\sin x}{8 + 2\sin x}$ for all $x$
