Use the limit comparison test to determine if the series converges or diverges.
-$$\sum _ { n = 2 } ^ { \infty } \frac { 1 } { 5 + 8 n \ln ( \ln n ) }$$
A) Diverges
B) Converges

Question

Quizplus · Accepted Answer

The limit comparison test involves comparing the given series to a simpler series whose convergence or divergence is known. For the given series $\sum _ { n = 2 } ^ { \infty } \frac { 1 } { 5 + 8 n \ln ( \ln n ) }$, we can compare it to the series $\sum _ { n = 2 } ^ { \infty } \frac { 1 } { n }$ because as $n$ becomes very large, the $8n\ln(\ln n)$ term dominates the denominator, making the $5$ negligible, and the behavior of the series is similar to $\frac{1}{n}$. The series $\sum _ { n = 2 } ^ { \infty } \frac { 1 } { n }$ is a p-series with $p = 1$, which is known to diverge. Therefore, by the limit comparison test, if the limit of the ratio of the given series to the simpler series is positive and finite, the two series will either both converge or both diverge. Since the simpler series diverges, the given series also diverges.

Use the Limit Comparison Test to Determine If the Series ∑n=2∞15+8nln⁡(ln⁡n)\sum _ { n = 2 } ^ { \infty } \frac { 1 } { 5 + 8 n \ln ( \ln n ) }n=2∑∞​5+8nln(lnn)1​

Use the Limit Comparison Test to Determine If the Series $\sum _ { n = 2 } ^ { \infty } \frac { 1 } { 5 + 8 n \ln ( \ln n ) }$