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Simple analysis problem gone wrong

  1. Feb 15, 2005 #1
    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..... :bugeye: 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?
  2. jcsd
  3. Feb 15, 2005 #2


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

  4. Feb 15, 2005 #3
    sorry, this is a sequence.
  5. Feb 15, 2005 #4


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

  6. Feb 15, 2005 #5
    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?
  7. Feb 15, 2005 #6


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    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.

  8. Feb 15, 2005 #7

    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?
  9. Feb 15, 2005 #8
    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
  10. Feb 15, 2005 #9

    this is based on limit theorems...so I want to use these techniques...
  11. Feb 16, 2005 #10


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    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

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