Results 1 to 9 of 9

Math Help - Convergence of Sequence

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    21

    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= \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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    \sqrt {n^2  + 2n}  - n = \dfrac{{2n}}{{\sqrt {n^2  + 2n}  + n}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Plato View Post
    \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 \frac{\sqrt{n^2 + 2n} + n}{\sqrt{n^2 + 2n} + n}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2010
    Posts
    21
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2010
    Posts
    21
    Ok, I have divided the top and bottom of that limit by n, and I get this....

    \lim _{n\rightarrow \infty }2\, \left( {\frac {\sqrt {{n}^{2}+2\,n}}{n<br />
}}+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!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    \dfrac{{2n}}{{\sqrt {n^2  + 2n}  + n}} = \dfrac{2}{{\sqrt {1 + \frac{2}{n}~}  + 1}}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Aug 2010
    Posts
    21
    Got it Thanks!!!!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    I corrected that typo.

    That is simple first year algebra. Divide through by n.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sequence Convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 22nd 2010, 03:51 AM
  2. sequence convergence
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 26th 2009, 07:45 AM
  3. Replies: 7
    Last Post: October 12th 2009, 10:10 AM
  4. Replies: 6
    Last Post: October 1st 2009, 09:10 AM
  5. Replies: 6
    Last Post: October 24th 2008, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum