Find the sum of the infinite geometric series. $\sum _ { n = 1 } ^ { \infty } \left( - \frac { 1 } { 3 } \right) ^ { n }$
A) $\frac { 3 } { 2 }$
B) $\frac { 3 } { 4 }$
C) $- \frac { 1 } { 4 }$
D) $- \frac { 1 } { 2 }$
E) $- \frac { 3 } { 4 }$

Question

Quizplus · Accepted Answer

The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$, where $a$ is the first term and $r$ is the common ratio. Here, $a = \left(-\frac{1}{3}\right)^1 = -\frac{1}{3}$ and $r = -\frac{1}{3}$. Substituting these values gives $S = \frac{-\frac{1}{3}}{1 - (-\frac{1}{3})} = \frac{-\frac{1}{3}}{\frac{4}{3}} = -\frac{1}{4}$.

Find the Sum of the Infinite Geometric Series ∑n=1∞(−13)n\sum _ { n = 1 } ^ { \infty } \left( - \frac { 1 } { 3 } \right) ^ { n }∑n=1∞​(−31​)n

Find the Sum of the Infinite Geometric Series $\sum _ { n = 1 } ^ { \infty } \left( - \frac { 1 } { 3 } \right) ^ { n }$