# Thread: Limit Test Convergence Problem

1. ## Limit Test Convergence Problem

Does it converge or diverge? If it converges, then what does it converge to?

Series with n = 1 to infinity $\dfrac{n^{3} - 3}{n^{3} + n^{2} + 1}$

This one would use the limit test, but I can't find $b_{n}$ (The series to compare it to). Any strategies for finding it?

Limit Test is: Limit as n goes toward infinity of $\dfrac{a_{n}}{b_{n}}$

Rules:

If $b_{n}$ is infinity then $a_{n}$ diverges

If $b_{n} = 0$ then $a_{n}$ converges

If $b_{n}$ comes out to a value, then if $b_{n}$converges, then $a_{n}$ converges, and if $b_{n}$ diverges, then $a_{n}$ diverges.

2. ## Re: Limit Test Convergence Problem

No friends the limit comparison test doesn't go like that.
Also, the limit of the sequence is 1, not 0, so by Divergence Test the series diverges.

3. ## Re: Limit Test Convergence Problem

Limit comparison test is like this:
Given a(n), comparing it to b(n), both of which are positive.
The test is valid only if lim(n-> inf.) a(n)/b(n) gives you a non-zero finite value: b(n) converges then a(n) also converges; b(n) diverges then a(n) also diverges.

4. ## Re: Limit Test Convergence Problem

One thing you have to do is find out what $b_{n}$ should be. Therefore, you have to simplify the n terms. Correct? How would we go about this? The correct $b_{n}$ should be $\dfrac{1}{n}$ How do we get there using the arithmetic of exponents?

How is simplification and exponential arithmetic done on

$\dfrac{n^{3} - 3}{n^{3} + n^{2} + 1}$ ??

5. ## Re: Limit Test Convergence Problem

Are you required to do this one by limit test? The divergence test quickly shows it diverges

6. ## Re: Limit Test Convergence Problem

I understand that it diverges. Because it's compared to either $\dfrac{1}{n^{2}}$ or $\dfrac{1}{n}$, not sure which. However, there is a procedure of simplification to find $b_{n}$ that I'm confused about.

Originally Posted by Jonroberts74
Are you required to do this one by limit test? The divergence test quickly shows it diverges

7. ## Re: Limit Test Convergence Problem

Originally Posted by Jason76
I understand that it diverges. Because it's compared to either $\dfrac{1}{n^{2}}$ or $\dfrac{1}{n}$, not sure which. However, there is a procedure of simplification to find $b_{n}$ that I'm confused about.
NO! You are supposed to know that a series only has a chance of converging if the terms of the series vanish to 0. So if the limit of your terms is NOT 0, then the series can not converge!

\displaystyle \begin{align*} \frac{n^3 - 3}{n^3 + n^2 + 1} &= \frac{n^3 + n^2 + 1 - n^2 - 4}{n^3 + n^2 + 1} \\ &= \frac{n^3 + n^2 + 1}{n^3 + n^2 + 1} - \frac{n^2 + 4}{n^3 + n^2 + 1} \\ &= 1 - \frac{n^2 + 4}{n^3 + n^2 + 1} \\ &= 1 - \frac{\frac{1}{n} + \frac{4}{n^3}}{1 + \frac{1}{n} + \frac{1}{n^3}} \\ &\to 1 - 0 \textrm{ as } n \to \infty \\ &= 1 \end{align*}

The terms do NOT approach 0, so the series diverges!

8. ## Re: Limit Test Convergence Problem

Originally Posted by Prove It
NO! You are supposed to know that a series only has a chance of converging ...
You are correct in that one way to show that

$\Sigma\frac{n^3-3}{n^3+n^3+1}$
diverges is to show that $\frac{n^3-3}{n^3+n^3+1}$ does not go to zero as n goes to infinity. However, although it might be the easiest, it is not the only way. A comparison with the divergent series $\frac{1}{n}$ will also do quite nicely.

9. ## Re: Limit Test Convergence Problem

Originally Posted by Ishuda
You are correct in that one way to show that

$\Sigma\frac{n^3-3}{n^3+n^3+1}$
diverges is to show that $\frac{n^3-3}{n^3+n^3+1}$ does not go to zero as n goes to infinity. However, although it might be the easiest, it is not the only way. A comparison with the divergent series $\frac{1}{n}$ will also do quite nicely.
Which is completely stupid and pointless, considering that the FIRST thing anyone should do before testing the convergence of a series is to check the limit of the terms. This will tell you if you need to go any further. Work smart, not hard!