# Can't figure out how to evaluate a sequence as it goes to infinity.

1. Jan 30, 2012

### kripenwah

1. The problem statement, all variables and given/known data

An = (((-1)^(n-1))n)/(n^2 + 1)

I need to know if it converges or diverges and if it converges the limit.

2. Relevant equations

3. The attempt at a solution

I know it converges to 0. But I don't know how to show it when evaluating. I tried evaluation An| in the absolute value but I keep ending up with oo/oo

2. Jan 30, 2012

### Dick

Try dividing numerator and denominator by n.

3. Jan 30, 2012

### kripenwah

lin n -> oo abs (((n(-1)^(n-1))/(n + 1/n)) * (1/n)) So it ends up being (infinity/infinity) * 0 = 0. Is that really a valid way to evaluate it?

4. Jan 30, 2012

### Dick

No, it's not. But n*(1/n)=1. Take the absolute value and you've got 1/(n+1/n). What's that limit?

5. Jan 30, 2012

### kripenwah

The limit is to infinity.
http://www4d.wolframalpha.com/Calculate/MSP/MSP44701a03g21d31b6h65500003b8813i95da1dcbe?MSPStoreType=image/gif&s=25&w=119&h=40 [Broken]

Last edited by a moderator: May 5, 2017
6. Jan 30, 2012

### Dick

Why would you think that?

7. Jan 30, 2012

### kripenwah

Well the problem text says Determine whether the sequences converges or diverges. If it converges find the limit.

8. Jan 30, 2012

### Dick

I don't think limit n->infinity of 1/(n+1/n) is infinity.

9. Jan 30, 2012

### kripenwah

Not sure if I wrote the problem bad. Here is what the problem looks like the text. an=

10. Jan 30, 2012

### Dick

That's what I thought it was. I'm going to just have to repeat what I said to start with. Divide numerator and denominator by n. That doesn't change the limit. Now you don't have infinity/infinity. What form do you have?

11. Jan 30, 2012

### kripenwah

I don't see how it helps unless you take the absolute value of it.

12. Jan 30, 2012

### Dick

Doesn't the denominator go to infinity and the numerator not go to infinity? The numerator is bounded. Taking the absolute value wouldn't hurt. But it's not infinity/infinity anymore.

13. Jan 30, 2012

### kripenwah

Numerator will be -1 or 1. Divided by infinity it evaluates to 0. But I am not sure if that is a valid way to show that it evaluates to 0. In the book I am using (Stewart) it gives a therom if Lim n-> infinity |an = o than Lim n-> infinity |an = 0.

14. Jan 30, 2012

### Dick

I think you mean lim n->infinity |an|=0 then lim n->infinity an=0. |an|=1/(n+1/n). Surely that approaches 0. It has the form 1/infinity doesn't it? You probably don't need a epsilon style proof here.