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How to calculate.

  1. Nov 23, 2007 #1
    [Solved] How to calculate.

    1. The problem statement, all variables and given/known data

    Determine the centre, radius, and Interval of convergence of the power series.

    [tex] \sum_{n=1}^\infty \frac{1}{n} \frac{(x+2)}{2}^n [/tex]


    2. Relevant equations

    Radius of convergence

    [tex] \frac{1}{R} = L = \lim_{n \to {\infty}} \frac{a_n+1}{a_n} [/tex]

    so R = 1/L

    and the interval is (x-R,x+R)


    3. The attempt at a solution

    Finding the centre is easy. x = -2

    But the algebra stuff is getting me down. This is where I get confused. I took this from the solution manual and I can't figure out this result.

    [tex] \sum_{n=1}^\infty \frac{1}{n} \frac{(x+2)}{2}^n [/tex]

    So we have [tex] R = \lim \frac{2^{n+1}(n+1)}{2^n n} = 2 [/tex]
     
    Last edited: Nov 23, 2007
  2. jcsd
  3. Nov 23, 2007 #2

    HallsofIvy

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    Put { } around the "n+1" subscript.

    Did you copy both the original problem and the solution correctly? In your original formula, that "2" in the denominator was NOT to the n power and so you will NOT have 2n in your ratio. In fact the factors of "2" in both an+1[/sup] and an will cancel and play no part. It should be obvious that
    [tex]\sum_{n=1}^\infty \frac{1}{n}\frac{(x+2)^n}{2}[/tex]
    is the same as
    [tex]\frac{1}{2}\sum_{n=1}^\infty \frac{1}{n}\(x+2)^n[/tex]
    so the "1/2" plays no part in the convergence of the series.

    You are given that [itex]a_n= (1/n)(x+2)^n[/itex] so [itex]a_{n+1}= (1/(n+1))(x+2)^{n+1}[/itex]. According to the "ratio test", that will converge as long as, in the limit, the ratio of consectutive terms goes to a limit of less than 1. The ratio you want is
    [tex]\frac{1}{n+1}\frac{|x+2|^{n+1}}{2}\frac{n}{1}\frac{2}{|x+2|^n}[/tex]
    which reduces to
    [tex]\frac{n}{n+1}|x+2|}[/tex]
    What is the limit of n/(n+1)?
     
  4. Nov 23, 2007 #3
    Ok, first of all, thanks for taking time looking at this.

    Limit of n/(n+1) is 1

    This is exactly what I got. But how is R = 2 then?

    Is it right to say that the limit of [tex]\frac{n}{n+1}|x+2|}[/tex] is 2 ?

    I dont know what to do with the x in all these calculations.
     
  5. Nov 23, 2007 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Now, the limit of [tex]\frac{n}{n+1}|x+2|[/tex] is |x+2|. That will be less than 1 as long as -1< x+2< 1 or =-3< x< -1. The radius of convergence is 1.

    That was why I asked if you were sure you had copied everything correctly.
    If the problem were, instead,
    [tex]\sum_{n=1}^\infty \frac{1}{n}\left(\frac{x+2}{2}\right)^n[/tex]
    or (same thing)
    [tex]\sum_{n=1}^\infty \frac{1}{n}\frac{(x+2)^n}{2^n}[/itex]
    then the radius of convergence would be 2.
     
  6. Nov 23, 2007 #5
    You are right! I didn't copy the example right!

    It is indeed [tex] \sum_{n=1}^\infty \frac{1}{n}\left(\frac{x+2}{2}\right)^n[/tex]

    And then of course the radius of convergence is 2 !!

    Well, thanks alot HallsofIvy. I'm kinda happy though that I wasn't doing anything wrong in my calculations, just a "small" typo.

    Again thanks.
     
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