Answer
Converges to $\dfrac{1}{1-\sin x}$ for $x\ne (2k+1)\pi/2$ (where k is an integer)
Work Step by Step
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}$
We have a convergent geometric series with first term, $a=1$ and common ratio $r =\sin x$
$S=\dfrac{1}{1-\sin x}$
Hence, the given series converges to $\dfrac{1}{1-\sin x}$ for $x\ne (2k+1)\pi/2$ (where k is an integer)