1. ## 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?

2. 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

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

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

3. 1 + 0 + 0 - 125/5 = -24 ?

4. What are you on about? You're subbing 1 here and 0 there. Take the limit as n goes to infinity

5. Originally Posted by Mathstud28
It diverges because this is equivalent to

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

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?

6. 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.

7. 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?
$\frac{n^3}{n+5} \geq \frac{n^3}{n+5n}=\tfrac{1}{6}n^2$ and $\lim ~ \tfrac{1}{6}n^2 = \infty$

8. Got it. Thanks!