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Thread: Convergence of Sequence

  1. #1
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    Convergence of Sequence

    I need help figuring out how the convergence of a sequence.

    I have the answer, but I'm not understanding how it is calculated.

    It is the sequence,

    an=$\displaystyle \sqrt{n^2+2n}-n$

    I have the answer that the limit as n goes to infinity is 1. How would I calculate this limit for the sequence?
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  2. #2
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    $\displaystyle \sqrt {n^2 + 2n} - n = \dfrac{{2n}}{{\sqrt {n^2 + 2n} + n}}$
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  3. #3
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    Quote Originally Posted by Plato View Post
    $\displaystyle \sqrt {n^2 + 2n} - n = \dfrac{{2n}}{{\sqrt {n^2 + 2n} + n}}$
    I just want to add that Plato probably got this result by using the standard technique of multiplying the original expression by $\displaystyle \displaystyle \frac{\sqrt{n^2 + 2n} + n}{\sqrt{n^2 + 2n} + n}$.
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  4. #4
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    I still don't understand where to go from here. I've tried using L'Hopital's rule here, but that doesn't seem to work either. Can you show how the limit as n goes to infinity is 1?
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  5. #5
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    Quote Originally Posted by RU2010 View Post
    I still don't understand where to go from here. I've tried using L'Hopital's rule here, but that doesn't seem to work either. Can you show how the limit as n goes to infinity is 1?
    Divide by n the numerator and denominator of the expression in post #2 (and which you ought to derive using the suggestion in post #3). Note that 1/n --> 0 etc.
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  6. #6
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    Ok, I have divided the top and bottom of that limit by n, and I get this....

    $\displaystyle \lim _{n\rightarrow \infty }2\, \left( {\frac {\sqrt {{n}^{2}+2\,n}}{n
    }}+1 \right) ^{-1}$


    How do I simplify the radical part on the bottom? I should end up with 2/2 so that the entire limit is 1. But I'm stuck on simplifying the radical.

    Thanks!
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  7. #7
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    $\displaystyle \dfrac{{2n}}{{\sqrt {n^2 + 2n} + n}} = \dfrac{2}{{\sqrt {1 + \frac{2}{n}~} + 1}}$
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  8. #8
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    Got it Thanks!!!!
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  9. #9
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    I corrected that typo.

    That is simple first year algebra. Divide through by $\displaystyle n$.
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