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Not sure how to evaluate the sum for series

  1. Jul 28, 2008 #1

    XSK

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

    [​IMG]

    2. Relevant equations

    --

    3. The attempt at a solution

    really what has thrown me here is the power of n. as we know it converges at some x, i figured i dont need to do any convergence tests. i thought about it and can only come to the conclusion that it converges at x=0, as 1^anything is always 1. but the question says values, and i feel i am missing some part of the puzzle here. also i'm not sure how to evaluate the sum as my knowledge of series is very limited.

    thanks.
     
  2. jcsd
  3. Jul 28, 2008 #2
    Re: convergence

    I should follow your example, write up the solution neatly and scan it.

    Here's my take as follows: The series starts with n=0, which means the first value will be [tex]x^2[/tex] because anything to the zeroth power is 1.

    But hold on a second. The first term is x^2. You see an x^2 right there. That means we're dealing with a geometric series.

    Think you can take it from there, or do you need another hint?
     
  4. Jul 28, 2008 #3
    Re: convergence

    I fail to see how you arrive at that conclusion based on the fact that the first term of the series is x2.

    On the other hand thinking about a geometric series is the correct way to approach this problem.
     
  5. Jul 28, 2008 #4

    Dick

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    Re: convergence

    Absolutely right. The first term is x^2. What's the common ratio?
     
  6. Jul 29, 2008 #5

    XSK

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    Re: convergence

    erm ok, i figured that the ratio is 1/(1+x)

    then i subtracted the series times the ratio from the series and got s = x(1+x)

    is this what i'm supposed to do?
     
  7. Jul 29, 2008 #6

    HallsofIvy

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    Re: convergence

    I have no idea what you mean by "I subtracted the series times the ratio from the series".

    What is the sum of the geometric series
    [tex]\sum_{n=0}^\infty A r^n [/tex]?
     
  8. Jul 29, 2008 #7
    Re: convergence

    I have just started studying this topic, but couldn't the root test be of use here? My thoughts:

    [tex]

    \lim_{n\rightarrow\infty}\sqrt[n]{\left|\frac{x^2}{(1+x)^n}\right|}<1

    [/tex]

    Is the criterion for convergence. If we work this out:

    [tex]

    \lim_{n\rightarrow\infty}\left|\frac{x^{\frac{2}{n}}}{1+x}\right|<1

    [/tex]

    [tex]

    \frac{1}{|1+x|}<1

    [/tex]

    [tex] |x|>0 [/tex]

    But we see that the sum is 0 for [tex]x=0[/tex], therefore the series converge for all [tex]x\in R[/tex]
     
  9. Jul 29, 2008 #8

    Dick

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    Re: convergence

    You're application of the root test is fine. The solution of the inequality 1/|1+x|<1 is, uh, incomplete. There are negative values that work as well. But XSK should probably solve this using knowledge of the geometric series, since the sum is needed as well as the region of convergence.
     
  10. Jul 29, 2008 #9
    Re: convergence

    By [tex]|x|>0[/tex] I mean that [tex]x[/tex] can be any negative value. Or is the solution then incorrect as well?
     
  11. Jul 29, 2008 #10

    Dick

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    Re: convergence

    Ah, I read that as x>0. No, not all negative numbers work. Just some of them. Maybe we should let XSK work out the details, yes?
     
  12. Jul 30, 2008 #11

    HallsofIvy

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    Re: convergence

    Why not just do what was suggested before.

    The geometric series
    [tex]\Sum_{n= 0}^\infty Ar^n[/tex]
    converges for |r|< 1 and converges to
    [tex]\frac{A}{1- r}[/itex]

    Here, A= x^2 and r= 1+ x. What does that tell you?

    (I find it difficult to believe that you are dealing with series of functions but have never seen geometric series before.)
     
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