Let $a_n=\frac{n^2}{n^4-4}$ and $b_n=\frac{1}{n^2}.$

Calculate the following $limit.$

$\underset{n}{lim}\frac{a_n}{b_n}$

$(Givean$ exact answer. Use symbolic notation and fractions where needed. Enter DNE $if$ the $limit$ does not exist.$)$

$\underset{n}{lim}\frac{a_n}{b_n}=$

Determine the convergence $of{\displaystyle \underset{n=1}{\overset{}{}}}{a}_n.$

${\displaystyle \underset{n=1}{\overset{}{}}}{a}_n$ converges $by$ the Limit Comparison Test because ${\displaystyle \underset{n=1}{\overset{}{}}}{b}_n$ converges and $\underset{n}{lim}\frac{a_n}{b_n}$ exists and $is$ finite.

${\displaystyle \underset{n=1}{\overset{}{}}}{a}_n$ converges $by$ the Limit Comparison Test because ${\displaystyle \underset{n=1}{\overset{}{}}}{b}_n$ converges and $\underset{n}{lim}\frac{a_n}{b_n}$ does not exist.

${\displaystyle \underset{n=1}{\overset{}{}}}{a}_n$ diverges $by$ the Limit Comparison Test because ${\displaystyle \underset{n=1}{\overset{}{}}}{b}_n$ diverges and $\underset{n}{lim}\frac{a_n}{b_n}is$ infinite.

${\displaystyle \underset{n=1}{\overset{}{}}}{a}_n$ diverges $by$ the Limit Comparison Test because ${\displaystyle \underset{n=1}{\overset{}{}}}{b}_n$ diverges.