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

[kripenwah]

Homework Statement

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

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

Homework Equations

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 A_n| in the absolute value but I keep ending up with oo/oo

 

[Dick]

Try dividing numerator and denominator by n.

 

[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?

 

[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?

 

[kripenwah]

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

 

[Dick]

The limit is to infinity.

Why would you think that?

 

[kripenwah]

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

 

[Dick]

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

 

[kripenwah]

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

 

[Dick]

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

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?

 

[kripenwah]

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

 

[Dick]

I don't see how it helps.

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.

 

[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 |a_n = o than Lim n-> infinity |a_n = 0.

 

[Dick]

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 |a_n = o than Lim n-> infinity |a_n = 0.

I think you mean lim n->infinity |a_n|=0 then lim n->infinity a_n=0. |a_n|=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.