Let $a_n=\frac{2n-cos(n)}{n^3}$ and $b_n=\frac{1}{n^2}.$ Calculate the following limit$.$

$($Express numbers in exact form. Use symbolic notation and fractions where needed. Enter DNE if the limit does not exist.$)$

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

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

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

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

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