Use the limit comparison test to determine if the series converges or diverges.
-$$\sum _ { n = 1 } ^ { \infty } \frac { 1 } { 9 \sqrt { n } + 7 ( \ln n ) ^ { 2 } }$$
A) Converges
B) Diverges

Question

Quizplus · Accepted Answer

The limit comparison test involves comparing the given series to a simpler series that we know converges or diverges. For the series $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { 9 \sqrt { n } + 7 ( \ln n ) ^ { 2 } }$, as $n$ approaches infinity, the dominant term in the denominator is $7 ( \ln n ) ^ { 2 }$ because the logarithmic term grows faster than the square root term, but still slower than any polynomial term of $n$. Thus, we compare it to the series $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n }$, which is a p-series with $p = 1$, known to diverge.To formally apply the limit comparison test, we take the limit of the ratio of the given series to the simpler series as $n$ approaches infinity. If this limit is finite and positive, the two series either both converge or both diverge.$$\lim _ { n \to \infty } \frac { \frac { 1 } { 9 \sqrt { n } + 7 ( \ln n ) ^ { 2 } } } { \frac { 1 } { n } } = \lim _ { n \to \infty } \frac { n } { 9 \sqrt { n } + 7 ( \ln n ) ^ { 2 } }$$This limit does not evaluate to zero because the numerator grows faster than the denominator. However, the exact value of the limit is not as important as the fact that it indicates the given series behaves similarly to the harmonic series $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n }$ in terms of convergence. Since the harmonic series diverges, the given series also diverges.

Use the Limit Comparison Test to Determine If the Series ∑n=1∞19n+7(ln⁡n)2\sum _ { n = 1 } ^ { \infty } \frac { 1 } { 9 \sqrt { n } + 7 ( \ln n ) ^ { 2 } }n=1∑∞​9n​+7(lnn)21​

Use the Limit Comparison Test to Determine If the Series $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { 9 \sqrt { n } + 7 ( \ln n ) ^ { 2 } }$