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Homework Help: Infinite series

  1. Nov 4, 2004 #1
    Today we started on infinite series, I'm getting the material just fine and able to do most of the problems, but one is giving me problems.

    [tex]\lim_{n\to{a}} 2n/\sqrt{n^2+1}[/tex]

    I recognized that [tex]\infty/\infty[/tex] so I can use L'Hopital's rule. So taking the derivative of the numerator and denominator I get.

    [tex]\frac{d}{dn} 2n = 2[/tex]
    and
    [tex]\frac{d}{dn} \sqrt{n^2+1} = \frac{n}{\sqrt{n^2+1}}[/tex]

    Somehow the solutions manual is getting
    [tex]2/\sqrt{1+(1/n^2)}[/tex]

    and a final answer of 2.

    I can't see how they turned what I get for the derivative in the denominator into what they use. L'Hopital's rule twice would get rid of the n on top and put a 2 there, but that wouldn't change the square root.

    thanks for any help
     
    Last edited: Nov 4, 2004
  2. jcsd
  3. Nov 4, 2004 #2

    Galileo

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    Simply divide the numerator and the denominator by n. You get:
    [tex]\frac{2n}{\sqrt{n^2+1}}=\frac{2}{\sqrt{1+1/n^2}}[/tex]
    I'd advise against using l'hospitals rule. Although it may be valid to use in some cases, you generally have to show it actually is.
    In this case, n is restricted to integer values so the derivative doesn't exist.
     
  4. Nov 4, 2004 #3
    Thanks for the help,

    Could you go through a little of the algebra in the denominator, it's still not making sense to me how it's actually done. What property allows you to divide the radical by n and come up with that?
     
  5. Nov 4, 2004 #4
    Doh, nevermind, it popped into my head as I was taking a shower. Thanks again.
     
  6. Nov 4, 2004 #5

    HallsofIvy

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    By the way, in mathematics, there is a technical difference between "sequences" and "series". What you are dealing with here are "series", not "sequences".
     
  7. Nov 4, 2004 #6

    Galileo

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    Haha, not the other way around? :tongue2:
    Don't confuse them indeed. :wink:
     
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