Find the sum of the series.
-$$\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n - 1 } \frac { 9 } { 7 ^ { n } }$$
A) $\frac { 3 } { 2 }$
B) $\frac { 63 } { 8 }$
C) $\frac { 9 } { 8 }$
D) $\frac { 21 } { 2 }$

Question

Quizplus · Accepted Answer

We can rewrite the series as follows:

\begin{align*}
\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n - 1 } \frac { 9 } { 7 ^ { n } } &= \frac{9}{7^1} - \frac{9}{7^2} + \frac{9}{7^3} - \frac{9}{7^4} + \cdots \
&= 9\left(\frac{1}{7} - \frac{1}{7^2} + \frac{1}{7^3} - \frac{1}{7^4} + \cdotsight) \
&= 9\cdot \frac{1/7}{1+1/7} \
&= \frac{9}{8}.
\end{align*}
Hence, the answer is $\boxed{	extbf{(C)}}.$

Find the Sum of the Series ∑n=1∞(−1)n−197n\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n - 1 } \frac { 9 } { 7 ^ { n } }n=1∑∞​(−1)n−17n9​

Find the Sum of the Series $\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n - 1 } \frac { 9 } { 7 ^ { n } }$