Convergence Tests: Limit Comparison & Comparison Explained

[roam]

Homework Statement

Here is a worked example:

http://img79.imageshack.us/img79/1909/88760168.gif

I don't understand why they used the "limit comparison test" and not the "comparison test".

Why can't we just conclude that the series $$\sum^{\infty}_{n=1} (\frac{n+1}{n^2 +1})^3$$ converges by comparison with $$\frac{1}{n^3}$$?

What's the necessity of using the limit comparison test?

Furthermore, I would like to know in which cases it is more suitable to use the comparison test and when to use the limit comparison test. I appreciate some guidance here. Thanks.

 

[Mark44]

To use the comparison test you would have to show that ((n + 1)/(n^2 + 1))^3 <= 1/n^3, for all n larger than some value N_0. It might not be very easy to establish this inequality. The limit comparison gets around this problem.

 

[roam]

How did you first inspect that it would not be very easy to establish that $$(\frac{n+1}{n^2 +1})^3 \leq \frac{1}{n^3}$$?

 

[Billy Bob]

If I were using the regular comparison test on this, I would say

$$n+1\le n+n= 2n$$

and

$$n^2+1\ge n^2$$

so that

$$\frac{n+1}{n^2+1}\le\frac{2n}{n^2}=\frac{2}{n}$$

and thus

$$\left(\frac{n+1}{n^2+1}\right)^3\le\frac{8}{n^3}$$

 

[Office_Shredder]

Without context I assume the purpose of the example is to demonstrate the limit comparison test, not demonstrate how to prove the convergence of that series

 

[roam]

If I were using the regular comparison test on this, I would say

$$n+1\le n+n= 2n$$

and

$$n^2+1\ge n^2$$

so that

$$\frac{n+1}{n^2+1}\le\frac{2n}{n^2}=\frac{2}{n}$$

and thus

$$\left(\frac{n+1}{n^2+1}\right)^3\le\frac{8}{n^3}$$

Hi!
I can't follow exactly what you've done here, could you please explain a little more?

 

[Mark44]

If you want to show that, for example, a/b < c/d, the best way to do this is to show that a < c and b > d.

You want the numerator on the left to be smaller than the one on the right, and you want the denominator on the left to be bigger than the one on the right. A larger denominator is associated with a smaller value.

I assume that that was the part of Billy Bob's work that you didn't understand.

 

[roam]

I get it now, thank you.