## University Calculus: Early Transcendentals (3rd Edition)

Converges to $\dfrac{1}{1-\sin x}$ for $x\ne (2k+1)\pi/2$ (where k is an integer)
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)