Find the sum of the convergent series
$\sum _ { n = 0 } ^ { \infty } \left( \frac { 1 } { 6 ^ { n } } - \frac { 1 } { 7 ^ { n } } \right)$
A) $\frac { 1 } { 7 }$
B) $\frac { 1 } { 30 }$
C) $\frac { 1 } { 37 }$
D) $\frac { 1 } { 60 }$
E) $\frac { 1 } { 14 }$

Question

Find the sum of the convergent series
$\sum _ { n = 0 } ^ { \infty } \left( \frac { 1 } { 6 ^ { n } } - \frac { 1 } { 7 ^ { n } } \right)$ 
A) $\frac { 1 } { 7 }$
B) $\frac { 1 } { 30 }$
C) $\frac { 1 } { 37 }$
D) $\frac { 1 } { 60 }$
E) $\frac { 1 } { 14 }$

Quizplus · Accepted Answer

The given series can be split into two separate geometric series: $\sum _ { n = 0 } ^ { \infty } \frac { 1 } { 6 ^ { n } }$ and $-\sum _ { n = 0 } ^ { \infty } \frac { 1 } { 7 ^ { n } }$. The sum of a geometric series $\sum _ { n = 0 } ^ { \infty } ar^n$ is $\frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio. For the first series, $a = 1$ and $r = \frac{1}{6}$, giving a sum of $\frac{1}{1-\frac{1}{6}} = \frac{6}{5}$. For the second series, $a = 1$ and $r = \frac{1}{7}$, giving a sum of $\frac{1}{1-\frac{1}{7}} = \frac{7}{6}$. Subtracting the second sum from the first gives $\frac{6}{5} - \frac{7}{6} = \frac{36}{30} - \frac{35}{30} = \frac{1}{30}$.

Find the Sum of the Convergent Series ∑n=0∞(16n−17n)\sum _ { n = 0 } ^ { \infty } \left( \frac { 1 } { 6 ^ { n } } - \frac { 1 } { 7 ^ { n } } \right)∑n=0∞​(6n1​−7n1​)

Find the Sum of the Convergent Series $\sum _ { n = 0 } ^ { \infty } \left( \frac { 1 } { 6 ^ { n } } - \frac { 1 } { 7 ^ { n } } \right)$