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Converging Sequence

  1. May 9, 2007 #1
    1. The problem statement, all variables and given/known data

    Determine whether each of the sequences converges or diverges. If it converges, find its limit. Explain with sufficient details each claim.

    2. Relevant equations
    (k) [tex]a_n=\sqrt{n^2+1}-n[/tex]

    The Squeeze Theorem (If [tex]b_n<a_n<c_n[/tex], and both [tex]\displaystyle\lim_{x\rightarrow\infty}a_n, c_n = L[/tex], then [tex]\displaystyle\lim_{x\rightarrow\infty}b_n=L[/tex])

    3. The attempt at a solution

    I had the choice of deciding between the squeeze theorem and L'Hopital. L'Hopital seemed useless because I don't see a way to get this into the form of a quotient. I know the sequence is convergent to 0. Figured it would be best to try the Squeeze theorem, as it's really my only other option.

    I know that [tex]0 \leq \sqrt{n^2+1}-n[/tex], but after struggling for a while, I still have no idea how to get a larger limit on the other side equal zero (without arbitrarily choosing one, 1/x for instance works, but there's no logical way to get from my function to 1/x...) I know that [tex]n+1>\sqrt{n^2+1}[/tex], but the result is n+1-n=1, and the limit of 1 is 1, period. Likewise, I can get rid of the 1, but I'll have n left over, and the limit of n is infinity! Got any good tips on getting the other side of the squeeze theorem inequality?
  2. jcsd
  3. May 9, 2007 #2


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    Multiply by [tex]\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}[/tex] and watch the magic.
  4. May 9, 2007 #3


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    You stated the Squeeze theorem incorrectly, but I suppose you know the correct statement. I don't know what you mean by there being "no logical way" to get 1/n, but if 1/n works, then it works, so what's the problem? 1/n converges to 0 as n goes to infinity, if you can prove that [itex]\sqrt{n^2+1}-n\leq n^{-1}[/itex] for all n, then you're done. There's an easier way in my opinion, but do it your way first, and once you have your own answer I can give you mine.
  5. May 10, 2007 #4
    Ah thx quasar; I shoulda thought of rationalization. I need to remember that everytime I see a square root...

    When I meant "logical way" AKG, I meant that, just by looking at the equation, we can't conclude that 1/x is bigger always. I had to look at a calculator. I have to justify my choice for the boundaries of the squeeze theorem. However, quasar pointed out I could just rationalize the function and end up finding out that 1/n is bigger.
  6. May 10, 2007 #5


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    No, but you should be able to prove it.
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