Let ${\displaystyle \underset{n=1}{\overset{}{}}}{a}_{n}be$ the series you identified above and ${\displaystyle \underset{n=1}{\overset{}{}}}{b}_n={\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{n}.$ Calculate the $limit.$

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

$(Express$ numbers $in$ exact form. Use symbolic notation and fractions where needed.$)$

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

Does the original series converge?

${\displaystyle \underset{n=1}{\overset{}{}}}{a}_n$ diverges because $\underset{n}{lim}\frac{a_n}{b_n}is$ finite and ${\displaystyle \underset{n=1}{\overset{}{}}}{b}_n$ diverges.

${\displaystyle \underset{n=1}{\overset{}{}}}{a}_n$ converges because $\underset{n}{lim}\frac{a_n}{b_n}is$ finite and ${\displaystyle \underset{n=1}{\overset{}{}}}{b}_n$ diverges.

${\displaystyle \underset{n=1}{\overset{}{}}}{a}_n$ converges because $\underset{n}{lim}\frac{a_n}{b_n}is$ finite and ${\displaystyle \underset{n=1}{\overset{}{}}}{b}_n$ converges.

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