Use the direct comparison lest to determine whether ${\displaystyle \underset{n=1}{\overset{}{}}}\frac{n}{\sqrt[\phantom{2}]{2n^4-1}}$ converges or diverges

Select one:

a

comerges by compartion with ${\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{\sqrt[\phantom{2}]{2}}\frac{1}{n^2}$

b$.$ Converges by comparison with ${\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{\sqrt[\phantom{2}]{2}}\frac{1}{n^3}$

Diverges by comparion wan ${\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{\sqrt[\phantom{2}]{2}}\frac{1}{\sqrt[\phantom{2}]{n}}$

Diverges by companson with ${\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{\sqrt[\phantom{2}]{2}}\frac{1}{n}$