Use the limit comparison test to determine if the series converges or diverges.
-$$\sum _ { n = 1 } ^ { \infty } \frac { 3 } { 9 + 8 n ( \ln n ) ^ { 2 } }$$
A) Converges
B) Diverges

Question

Quizplus · Accepted Answer

We can compare the given series with the series $\sum_{n=1}^\infty \frac{1}{n(\ln n)^2}$, which is a known convergent series. Since both series have positive terms, we can use the limit comparison test:
$$\lim_{n\to\infty} \frac{\frac{3}{9+8n(\ln n)^2}}{\frac{1}{n(\ln n)^2}} = \lim_{n\to\infty} \frac{3n(\ln n)^2}{9+8n(\ln n)^2} = \frac{3}{8}$$
Since this limit is a nonzero finite number, and the series $\sum_{n=1}^\infty \frac{1}{n(\ln n)^2}$ converges, we can conclude that the given series also converges. Therefore, the best choice is A.

Use the Limit Comparison Test to Determine If the Series ∑n=1∞39+8n(ln⁡n)2\sum _ { n = 1 } ^ { \infty } \frac { 3 } { 9 + 8 n ( \ln n ) ^ { 2 } }n=1∑∞​9+8n(lnn)23​

Use the Limit Comparison Test to Determine If the Series $\sum _ { n = 1 } ^ { \infty } \frac { 3 } { 9 + 8 n ( \ln n ) ^ { 2 } }$