Use the limit comparison test to determine if the series converges or diverges.
-$$\sum _ { n = 1 } ^ { \infty } \frac { 5 \sqrt { n } } { 6 n ^ { 3 / 2 } + 1 n + 2 }$$
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 $\sum _ { n = 1 } ^ { \infty } \frac { 5 \sqrt { n } } { 6 n ^ { 3 / 2 } + n + 2 }$, we compare it to the series $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n }$ by examining the limit of their ratio as $n$ approaches infinity. The dominant term in the numerator is $5\sqrt{n}$ and in the denominator is $6n^{3/2}$. Simplifying, we get $\lim _ { n \rightarrow \infty } \frac { 5 \sqrt { n } } { 6 n ^ { 3 / 2 } + n + 2 } \cdot \frac { n } { 1 } = \lim _ { n \rightarrow \infty } \frac { 5n } { 6n^{3/2} + n^{3/2} + 2n^{1/2} } = \lim _ { n \rightarrow \infty } \frac { 5 } { 6n^{1/2} + 1 + 2/n^{1/2} }$, which approaches 0. However, the comparison is with $\frac{1}{n}$, which diverges (harmonic series). Since the limit of the ratio of the given series to a divergent series is finite (and not zero), the given series also diverges by the limit comparison test.

Use the Limit Comparison Test to Determine If the Series ∑n=1∞5n6n3/2+1n+2\sum _ { n = 1 } ^ { \infty } \frac { 5 \sqrt { n } } { 6 n ^ { 3 / 2 } + 1 n + 2 }n=1∑∞​6n3/2+1n+25n​​

Use the Limit Comparison Test to Determine If the Series $\sum _ { n = 1 } ^ { \infty } \frac { 5 \sqrt { n } } { 6 n ^ { 3 / 2 } + 1 n + 2 }$