# Simple analysis problem gone wrong

1. Feb 15, 2005

### semidevil

man, so I asked some people about some questions, and they confuse the heck out of me, so i'm gonna stay here for a while.

anyways, all I want to do is to prove ((-1)^n)n/(n+1) converge or diverge. I claim it is divergent, but others claim that (-1)^n doesn't converge nor diverge..... confused.

anywys, can you guys check it for me.

so for my proof, I multiply top and bottom by 1/n, and I get (-1)^n/(1+ 1/n). taking the limit, I see that the bottom will tend to 1, and the top will just go from -1 to 1. so this becomes *something that diverges*/1. which means this divergres.....is this valid?

2. n^2 /(n+1). again, I multiply by 1/n, and get n/(1+1/n). again, the top will diverge, and the bottom will tend to 1. So is it ok to say that this will diverge because it is *something that diverges*/1?

3. (2n^2 + 3)/(n^2 + 1). same thing, multiply by 1/n, and get (2n + 3/n)/(n + 1/n). ok, so if I get stuck here. Can I take the limit of the 3/n and 1/n, and let them tend to 0. and then this becomes 2n + 0/ n + 0, which simplifies to 2.

2. Feb 15, 2005

### dextercioby

Are u talking about a sequence (of numbers) or about (the sum of) a series ...??

Daniel.

3. Feb 15, 2005

### semidevil

sorry, this is a sequence.

4. Feb 15, 2005

### dextercioby

Then it obviously diverges...And i hope u know why...

Daniel.

5. Feb 15, 2005

### semidevil

to me, I know that (-1)^n will always bounce from -1 to 1, so it wont converge to any limit.

now, the way that I proved these few statements......do they seem correct? did I jump to any conclusions?

6. Feb 15, 2005

### dextercioby

All three examples diverge.The second would diverge,even if the (-1)^{n} factor would be missing.The proofs you've offered are rather cumbersome.

Daniel.

7. Feb 15, 2005

### semidevil

ooh....that's really bad...I must be missing something really basic then...um....can you kind of give me advice on these proofs if they aren't sufficient?

8. Feb 15, 2005

### xorbie

If you want to start from first principles, let the limit equal L. Then just show that there exists an epsilon, such that that there is no N: n>N --> | f(n) - L | < epsilon

9. Feb 15, 2005

### semidevil

this is based on limit theorems...so I want to use these techniques...

10. Feb 16, 2005

### learningphysics

Yes it is essentially valid. You should use the theorem that says
limn->inf a_n/b_n = (limn->inf a_n )/(limn->inf b_n) and simplify

Eventually you should get limn->inf (-1)^n and at this point you can say, the sequence diverges. (Diverges means "not converge"... you can't have a sequence that doesn't converge or diverge)

Remember you can take constants out of the limit

Again use the theorem for division
limn->inf (n/(1+1/n))
(limn->inf n)/(limn->inf 1+1/n) =
(limn->inf n)/1=
lim n->inf n which diverges (I don't think you have to prove this part)

Yes, the above converges. I'd multiply numerator and denominator by 1/n^2