# Investigate convergence

1. Mar 21, 2009

### danielc

1. The problem statement, all variables and given/known data
Investigate convergence of $$\sum$$$$\frac{k+k^{1/2}}{k^{3}-4k+3}$$

3. The attempt at a solution
I am trying to use comparison test to investigate the convergence , but I am having trouble of finding a correct term to be compared to. please help
1. The problem statement, all variables and given/known data

3. The attempt at a solution

2. Mar 21, 2009

### Dick

Based on it's behavior at large k, do you think it converges or diverges and why? Your first step in finding a comparison is to figure out if you want to compare with something larger or smaller.

3. Mar 21, 2009

### danielc

Base on the behavior of k , I think the series will converges , so I would have to compare it with something larger (called it b) , thus , if b converges than the original series converges. However, I am having hard time figuring out what converging b I should choose which won't violate the property of original series. Could you give me some hint?

4. Mar 21, 2009

### Dick

Sure. If you have a/b and you want to create something larger, then you want to increase the numerator and decrease the denominator. I'll give you a big hint. How about (k^3)/2 for the denominator? Remember you don't have to worry about small values of k. The comparison only has to hold for large k. Now the numerator? Keep it simple.

Last edited: Mar 21, 2009
5. Mar 21, 2009

### danielc

If I make numerator to be k+k=2k instead of k+k^1/2 , does that work?
but is it okay to ignore the -4k term in the original denominator ?

6. Mar 21, 2009

### Dick

It works great. But we aren't really ignoring the -4k+3 in the denominator. We are just saying that for k large enough, that k^3-4k+3>(k^3)/2. I think in this case k>2. But it's not terribly important exactly where. You can always use (k^3)/100 if you want. The proof still works.

7. Mar 21, 2009

### danielc

thank you very much , I think I got the idea now .