Simple analysis problem gone wrong

[semidevil]

man, so I asked some people about some questions, and they confuse the heck out of me, so I'm going to 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.

but I feel bad about this, because I'm not taking the limit of everything at the same time...you know?

 

[dextercioby]

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

Daniel.

 

[semidevil]

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

Daniel.

sorry, this is a sequence.

 

[dextercioby]

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

Daniel.

 

[semidevil]

to me, I know that (-1)^n will always bounce from -1 to 1, so it won't converge to any limit.

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

 

[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.

 

[semidevil]

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.

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?

 

[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

 

[semidevil]

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

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

 

[learningphysics]

man, so I asked some people about some questions, and they confuse the heck out of me, so I'm going to 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?

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

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?

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)

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.

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

Your answers seem right. 1 and 2 diverge. 3 converges.