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Analysis: What does Sqrt(n^2+2n)-n converge to?Prove it.

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Formulate a conjecture about the convergence or divergence of the sequence:

    [itex]\sqrt{n^{2}+2n} - n[/itex]

    2. Relevant equations

    Triangle Inequality, etc...

    3. The attempt at a solution

    Start by noticing that [itex](n+1)^{2} = (n^{2}+2n+1), so, (n+1) = \sqrt{n^{2}+2n+1}[/itex]

    Now, [itex]\sqrt{n^{2}+2n} - n < |\sqrt{n^{2}+2n}-(n+1)|[/itex]

    Multiply the RHS by 1 in the following manner:

    [itex]|\sqrt{n^{2}+2n}-(n+1)| * \frac{\sqrt{n^{2}+2n+1}+\sqrt{n^{2}+2n}}{\sqrt{n^{2}+2n+1}+\sqrt{n^{2}+2n}}[/itex]

    [itex]= \frac{n^{2}+2n+1-n^{2}-2n}{\sqrt{n^{2}+2n}+(n+1)}[/itex]

    [itex]= \frac{1}{\sqrt{n^{2}+2n}+(n+1)}[/itex]

    Now, by observation,

    [itex]\frac{1}{\sqrt{n^{2}+2n}+(n+1)} < \frac{1}{n}[/itex]

    we know [itex]\forall\epsilon>0 \exists N\in N \ni n \geq N => \frac{1}{n} < \epsilon[/itex]

    I'm not sure what to do now...
     
    Last edited: Oct 4, 2011
  2. jcsd
  3. Oct 3, 2011 #2
    Also, could someone tell me why the latex formating didn't work in the above post...
     
  4. Oct 3, 2011 #3

    SammyS

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    Yes,

    You terminated them with a backslash [\itex] rather than a slash [/itex] .
     
  5. Oct 4, 2011 #4

    Dick

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    And I think you want to multiply numerator and denominator by the 'conjugate' sqrt(n^2+2n)+n. Then think about the limit.
     
  6. Oct 4, 2011 #5
    lol, stupid mistake,

    Thank you
     
  7. Oct 4, 2011 #6

    SammyS

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    A mistake... NOT a stupid mistake.
     
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