Use the Limit Comparison Test (if possible) to determine whether the series $\sum _ { n = 1 } ^ { \infty } \frac { 2 n ^ { 2 } - 7 } { 7 n ^ { 7 } + 3 n + 2 }$ converges or diverges.
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 $\sum _ { n = 1 } ^ { \infty } \frac { 2 n ^ { 2 } - 7 } { 7 n ^ { 7 } + 3 n + 2 }$, the dominant term in the numerator is $2n^2$ and in the denominator is $7n^7$. Comparing it to the series $\sum _ { n = 1 } ^ { \infty } \frac {1}{n^5}$ (since the powers of $n$ in the dominant terms give $n^2/n^7 = 1/n^5$), we find that the series $\frac {1}{n^5}$ converges (it is a p-series with $p > 1$). The limit of the ratio of the given series to the simpler series is a positive finite number, indicating that both series behave similarly in terms of convergence/divergence. Since $\sum _ { n = 1 } ^ { \infty } \frac {1}{n^5}$ converges, by the Limit Comparison Test, the given series also converges.

Use the Limit Comparison Test (If Possible) to Determine Whether ∑n=1∞2n2−77n7+3n+2\sum _ { n = 1 } ^ { \infty } \frac { 2 n ^ { 2 } - 7 } { 7 n ^ { 7 } + 3 n + 2 }∑n=1∞​7n7+3n+22n2−7​

Use the Limit Comparison Test (If Possible) to Determine Whether $\sum _ { n = 1 } ^ { \infty } \frac { 2 n ^ { 2 } - 7 } { 7 n ^ { 7 } + 3 n + 2 }$