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Power Series Converge Absolutely

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data
    for what values of x does the series converge absolutely?


    2. Relevant equations
    [itex]\displaystyle \sum^{∞}_{n=1} \frac{4^n * x^n}{n!}[/itex]


    3. The attempt at a solution
    Ratio Test

    [itex]\displaystyle \frac{4^{n+1} * x^{n+1}}{n+1)!} * \frac{n!}{4^n * x^n}[/itex]
    4x * limit (n->inf) [itex]\displaystyle \frac{1}{n+1} = 0[/itex]

    What do I do now since the limit is zero? I asked a similar question in another thread, but the limit turned out to be 1. I am sure that this limit is zero.
     
  2. jcsd
  3. Apr 4, 2013 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What was your purpose in calculating that? How is that number, whether it is 0 or 1 or whatever, tell you about the "radius of convergence"?
     
  4. Apr 4, 2013 #3

    Mark44

    Staff: Mentor

    It means that the series converges for all values of x.

    BTW, when you use the Ratio Test, you should be working with the absolute values of the terms in your series.

    This is the ratio you should be working with:
    $$ \frac{4^{n+1} * |x|^{n+1}}{(n+1)!} * \frac{n!}{4^n * |x|^n}$$
    The result you get will be 4|x| ## \lim_{n \to \infty} 1/(n + 1)## = 0, which places no limits on the values of x.
     
  5. Apr 4, 2013 #4
    do i leave the answer at 4|x| limn→∞1/(n+1) = 0 and put -inf < x < inf. or do i need to prove this somehow?
     
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