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Homework Help: Series expansion of xln((x+1)/x)

  1. Oct 9, 2011 #1
    basically i have to check if
    [tex] xln\frac{(x+1)}{x} [/tex]→ 1 as x→∞


    the first term is 0 as x→∞

    in the answers they say they used maclaurin series and got

    [tex]x(\frac{1}{x} + O\frac{1}{1^{2}}) [/tex]

    but dont show how they did it.

    would the first term in the series be

    [tex]a(ln(\frac{a+1}{a}))[/tex]

    there a = 0, but then the ln function is defined?

    im getting a little confused about have a=0 but we want to figure out where the function goes to as x→∞
     
  2. jcsd
  3. Oct 9, 2011 #2
    Looks to me it should be:

    [tex]1/x(x+O(k))[/tex]

    if instead of taking the limit at infinity, we take the limit at zero of the expression:

    [tex]1/x\ln\left(\frac{1+1/x}{1/x}\right)[/tex]

    Now we can express the log expression as a MacLaurin series.
     
  4. Oct 12, 2011 #3
    sorry i dont understand what you mean.

    are you rewriting our original expression?

    how can i write the ln(1+(1/x)) expression so that its defined for x=0
     
  5. Oct 12, 2011 #4

    SammyS

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    Why is jackmell's suggestion useful?

    [itex]\displaystyle \frac{1+1/x}{1/x}=x+1[/itex]
     
  6. Oct 12, 2011 #5

    Ray Vickson

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    You should not care what happens near x = 0; after all, the question is asking about what happens as x --> infinity. So, if x is large, y = 1/x is small, and surely you know how ln(1+y) behaves near y = 0.

    RGV
     
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