Use the Comparison Test to determine if the series converges or diverges.
-$$\sum _ { n = 1 } ^ { \infty } \frac { 1 } { 6 ^ { n - 1 } + 2 }$$
A) diverges
B) converges

Question

Quizplus · Accepted Answer

Since $6^{n-1}+2 > 6^{n-1}$ for all $n$, we have 
$$
0 < \frac{1}{6^{n-1}+2} < \frac{1}{6^{n-1}} = \frac{1}{6}\cdot\frac{1}{6^{n-2}}
$$
for all $n$.

Now, since $\sum_{n=1}^\infty \frac{1}{6^{n-2}}$ converges (it is a geometric series with ratio $\frac16<1$), by the Comparison Test, we have that 
$$
\sum_{n=1}^\infty \frac{1}{6^{n-1}+2}
$$
converges.

Use the Comparison Test to Determine If the Series Converges ∑n=1∞16n−1+2\sum _ { n = 1 } ^ { \infty } \frac { 1 } { 6 ^ { n - 1 } + 2 }n=1∑∞​6n−1+21​

Use the Comparison Test to Determine If the Series Converges $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { 6 ^ { n - 1 } + 2 }$