Comparison test to determine convergence

[Needassistance0987]

Homework Statement

Hello, is using a comparison test for this question ok? because it looked weird to me somehow

Relevant Equations

if bn>an
then an divergence bn also divergence

 

[Mark44]

Homework Statement: Hello, is using a comparison test for this question ok? because it looked weird to me somehow
Relevant Equations: if bn>an
then an divergence bn also divergence

View attachment 343263

To use the comparison test, you need to compare the general term in the series you're looking at with the general term of a series that is known to diverge. Have you convinced yourself that $\sum_{n=1}^\infty (\sqrt[n] 2 - 2)$ diverges? If so, then your series also diverges.

You could also use the Nth Term Test for Divergence. If $\lim_{n \to \infty} a_n \ne 0$, then the series $\sum a_n$ diverges.

Also, we don't say "$a_n$ divergence" -- we say that $\sum a_n$ diverges or it converges.

 

[FactChecker]

That will not work. To use the direct comparison test, you have to compare positive numbers. $\sqrt[n]{2} -2$ becomes negative. See this. You can use absolute values and compare it with an absolutely convergent series to prove absolute convergence. See this.