Use the Limit Comparison Test to determine the convergence or divergence of the series $\sum _ { n = 1 } ^ { \infty } \frac { 3 ^ { n } + 1 } { 8 ^ { n } + 1 }$.
A)
The series $\sum _ { n = 1 } ^ { \infty } \frac { 3 ^ { n } + 1 } { 8 ^ { n } + 1 }$ converges.
B)
The series $\sum _ { n = 1 } ^ { \infty } \frac { 3 ^ { n } + 1 } { 8 ^ { n } + 1 }$ diverges.

Question

Quizplus · Accepted Answer

We apply the limit comparison test with the series $\sum_{n=1}^\infty \frac{1}{8^n}$.

$$ L= \lim_{n \rightarrow \infty} \frac{\frac{3^n + 1}{8^n + 1}}{\frac{1}{8^n}} = \lim_{n \rightarrow \infty} \frac{3^n+1}{8^n+1} \cdot \frac{8^n}{1} \cdot \frac{1}{3^n} = \lim_{n \rightarrow \infty} \frac{8^n}{3^n} \cdot \frac{3^n+1}{8^n+1} = \lim_{n \rightarrow \infty} \frac{3^n+ \frac{1}{8^n}}{8^n+1} = 0 $$

Since $L<\infty$ and the series $\sum_{n=1}^\infty \frac{1}{8^n}$ converges, by the Limit Comparison Test we conclude that the series $\sum _ { n = 1 } ^ { \infty } \frac { 3 ^ { n } + 1 } { 8 ^ { n } + 1 }$ converges.

Use the Limit Comparison Test to Determine the Convergence or Divergence