# Converge or Diverge?

• Nov 3rd 2008, 04:25 PM
veronicak5678
Converge or Diverge?
An = n^3 / (n + 5)

Can I just take the limit and say this is equal to 1 / 0 so it diverges?
• Nov 3rd 2008, 04:52 PM
Mathstud28
Quote:

Originally Posted by veronicak5678
An = n^3 / (n + 5)

Can I just take the limit and say this is equal to 1 / 0 so it diverges?

It diverges because this is equivalent to

$\displaystyle \frac{n+125-125}{n+5}=\frac{n^3+125}-\frac{125}{n+5}=n^2+5n+25-\frac{125}{n+5}$

Now what is the limit of that as it goes to infinity?
• Nov 3rd 2008, 05:37 PM
veronicak5678
1 + 0 + 0 - 125/5 = -24 ?
• Nov 3rd 2008, 05:41 PM
Chop Suey
What are you on about? You're subbing 1 here and 0 there. Take the limit as n goes to infinity
• Nov 3rd 2008, 05:41 PM
Jameson
Quote:

Originally Posted by Mathstud28
It diverges because this is equivalent to

$\displaystyle \frac{n+125-125}{n+5}=\frac{n^3+125}-\frac{125}{n+5}=n^2+5n+25-\frac{125}{n+5}$

Now what is the limit of that as it goes to infinity?

While that's true, isn't is much easier to simply take the original form of A_n and easily see that the limit is infinity as n approaches infinity?
• Nov 3rd 2008, 05:46 PM
veronicak5678
I think that's more along the lines of what I was supposed to be doing, anyway...

So I'll just take the limit off the bat and say it diverges.
• Nov 3rd 2008, 05:48 PM
ThePerfectHacker
Quote:

Originally Posted by veronicak5678
An = n^3 / (n + 5)

Can I just take the limit and say this is equal to 1 / 0 so it diverges?

$\displaystyle \frac{n^3}{n+5} \geq \frac{n^3}{n+5n}=\tfrac{1}{6}n^2$ and $\displaystyle \lim ~ \tfrac{1}{6}n^2 = \infty$
• Nov 3rd 2008, 05:57 PM
veronicak5678
Got it. Thanks!