Question

Determine whether the series convergent or divergent (sqrt(n)+4)/n^2?

I'm at a loss for this one. Please do not just tell me "this series converges." I know it does.
Is this just a p-series test?

Answer 1

\[\Large\rm \sum_{n=1}^{\infty}\frac{\sqrt{n}+4}{n^2}\ge\sum_{n=1}^{\infty}\frac{\sqrt{n}}{n^2}\qquad \forall ~n\ge1\]Hmmm comparison test maybe?
Seems like this should work. Yes, relating it to P-series as you suggested.

Answer 2

\[0\le\frac{\sqrt{n}+4}{n^2}\le\frac{\sqrt{n}+4\sqrt{n}}{n^2}\le\frac{5\sqrt{n}}{n^2}\]

for \(n\ge 1\)

Answer 3

I guess I could have put

\[\frac{\sqrt{n}+4\sqrt{n}}{n^2}=\frac{5\sqrt{n}}{n^2}\]

but it is still technically true

Answer 4

Oh we want to show that our series is less than a convergent series :3
lol ya i was thinking backwards XD

Answer 5

I got it! Thank you very much!