Determine whether the sequence defined by $a _ { n } = \frac { n ^ { 2 } - 5 } { 6 n ^ { 2 } + 1 }$ converges or diverges. If it converges, find its limit. Select the correct answer.
A) $\frac { 1 } { 6 }$
B) $- 5$
C) $- \frac { 5 } { 6 }$
D) Diverges

Question

Quizplus · Accepted Answer

We can use limit comparison test to determine its convergence. Note that
$$\lim_{n\to\infty}\frac{n^2}{6n^2+1}= \lim_{n\to\infty}\frac{1}{6+\frac{1}{n^2}}=\frac{1}{6}$$and $\sum_{n=1}^\infty\frac{1}{n}$ is a divergent harmonic series. Therefore, by limit comparison test,
$$\lim_{n\to\infty}\frac{a_n}{\frac{1}{n}}=\lim_{n\to\infty}\frac{n^2-5}{6n^3+1}n\to\frac{1}{6}\cdot 1=\frac{1}{6}<\infty.$$
Thus, the series converges by comparison test, and the limit is 
$$\lim_{n\to\infty} a_n = \lim_{n\to\infty}\frac { n ^ { 2 } - 5 } { 6 n ^ { 2 } + 1 } = \frac { 1 } { 6 },$$
so the answer is A.

Determine Whether the Sequence Defined By an=n2−56n2+1a _ { n } = \frac { n ^ { 2 } - 5 } { 6 n ^ { 2 } + 1 }an​=6n2+1n2−5​

Determine Whether the Sequence Defined By $a _ { n } = \frac { n ^ { 2 } - 5 } { 6 n ^ { 2 } + 1 }$