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Power series

  1. Apr 24, 2007 #1
    please take a look at the attachment...

    i'm did the |an+1/an|
    and then i dont' know how to deal with ln
     

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  2. jcsd
  3. Apr 24, 2007 #2

    quasar987

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    You've got the fraction backward. I.e. your numerator should be your denominator and vice versa.

    In any case, you could try to break the thing with l'Hospital.
     
  4. Apr 24, 2007 #3

    HallsofIvy

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    I take it you are attempting to find the radius of convergence. Whether you have the fraction upside down or not depends on exactly "where" you are. Certainly to use the "ratio test", you would form
    [tex]\frac{(n+3)ln(n+3)}{(n+2)ln(n+2)}[/tex]
    but once you had that, your "radius of convergence" is the reciprocal of the limit of that- the limit of the fraction you have.

    Obviously the limit of (n+2)/(n+3) (or (n+3)/(n+2)) is 1 so everything depends on the log terms. Treat it as a continuous problem: ln(x+3)/ln(x+2) and use L'Hopital's rule. As it turns out, it really doesn't matter which way you have the fraction!
     
  5. Apr 24, 2007 #4

    quasar987

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    Are you sure? in the ratio test, we look at the limit of [tex]|a_{n+1}/a_n|[/tex], and the radius of convergence is the reciprocal of this. Here,

    [tex]a_n=\frac{1}{(n+2)\ln(n+2)}[/tex],

    so,

    [tex]\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{(n+2)\ln(n+2)}{(n+3)\ln(n+3)}[/tex]
     
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