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Homework Help: Some proofs of convergence

  1. Oct 16, 2008 #1
    1. The problem statement, all variables and given/known data
    1. Prove that the sequence sqrt(n+1) - sqrt(n) converges to 0.
    2. If sequence {an} is composed of real numbers and if lim as n goes to infinity of {a2n} = A and the limit as n goes to infinity of {a(2n-1)} = A, prove that {an} converges to 1. Is converse true?
    3. Consider sequences {an} and {bn}, where bn = (an)^(1/n)
    a. If {bn} converges to 1, does the sequence {an} necessarily converge?
    b. If {bn} converges to 1, does the sequence {an} necessarily diverge?
    c. does {bn} have to converge 1?

    2. Relevant equations

    3. The attempt at a solution
    I'm not sure if I can divide sqrt(n) by sqrt(n) and prove that this new sequence goes to 1 without a loss of generality. As for the others, I am new to these proofs and any help would be much appreciated.
  2. jcsd
  3. Oct 16, 2008 #2


    Staff: Mentor

    1. Certainly you can divide sqrt(n) by itself, as long as n is not 0, but why would you want to do this? Even if you did want to do this, it would be trivial to prove that the limit of that sequence {sqrt(n)/sqrt(n)} is 1.

    Instead, what about multiplying the numerator and denominator by sqrt(n+1) + sqrt(n)? You'd be multiplying by 1, so this won't change the value of the terms in the sequence.

    2. If all the even-subscript terms in the sequence are approaching A, and the odd-subscript terms are doing the same thing, you're going to have a difficult time proving the sequence converges to 1.
  4. Oct 17, 2008 #3


    User Avatar
    Science Advisor

    There is no reason to do that. As Mark44 said, multiply "numerator and denominator" by sqrt{n+1}+ sqrt{n}.

    For problem two are you sure it didn't say "prove that {an} converges to A"? That would make a lot more sense.
    Last edited by a moderator: Oct 23, 2008
  5. Oct 23, 2008 #4
    Wouldn't you just use the def of convergence to prove No. 1
  6. Oct 23, 2008 #5
    need a little help with Xn= (cos n)/(n^3-n^2) and what it converges to.
  7. Oct 23, 2008 #6


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    Science Advisor

    Do NOT add new problems to someone else's threads. Start your own thread.

    Here's a hint: [itex]-1\le cos(n)\le 1[/itex]. Of course, you are assuming n> 1.
  8. Oct 23, 2008 #7
    Sorry I am a newbie!!!
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