Find the sum of the geometric series for those x for which the series converges.
-$$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 7 } \frac { 1 } { ( 3 + \sin x ) ^ { n } }$$
A) $\frac { 3 + \sin x } { 7 ( 4 - \sin x ) }$
B) $\frac { 3 + \sin x } { 7 ( 4 + \sin x ) }$
C) $\frac { 7 + \sin x } { 3 ( 4 - \sin x ) }$
D) $\frac { 7 + \sin x } { 3 ( 4 + \sin x ) }$

Question

Quizplus · Accepted Answer

The given series is a geometric series with the first term $a = \frac{1}{7}$ (when $n = 0$) and common ratio $r = \frac{-1}{3 + \sin x}$. The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$, provided $|r| < 1$. Substituting $a$ and $r$ into this formula gives $S = \frac{\frac{1}{7}}{1 - \frac{-1}{3 + \sin x}} = \frac{1}{7} \cdot \frac{3 + \sin x}{4 + \sin x} = \frac{3 + \sin x}{7(4 + \sin x)}$, which matches option B.

Find the Sum of the Geometric Series for Those X ∑n=0∞(−1)n71(3+sin⁡x)n\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 7 } \frac { 1 } { ( 3 + \sin x ) ^ { n } }n=0∑∞​7(−1)n​(3+sinx)n1​

Find the Sum of the Geometric Series for Those X $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 7 } \frac { 1 } { ( 3 + \sin x ) ^ { n } }$