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Radius of Convergence of power series anx^n^2

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that the power series [itex]\sum[/itex]anxn for n=0 to n=∞ has a radius of convergence R[itex]\in[/itex](0,∞). Find the radii of convergence of the series [itex]\sum[/itex]anxn2 from n=0 to n=∞ and [itex]\sum[/itex]anx2n.


    2. Relevant equations

    Radius of convergence theorem:

    R = 1/limsup|an|1/n is the radius of convergence of the series [itex]\sum[/itex]an(x-c)n

    Root Test:

    Let L = limsup|an|1/n, then

    If L<1, the series [itex]\sum[/itex]an converges absolutely
    If L>1, the series diverges
    If L=1, then the test is inconclusive.


    3. The attempt at a solution

    So we know that limsup|an|1/n = [itex]\frac{1}{R}[/itex].

    So for the second series ([itex]\sum[/itex]anx2n) I think this is easy:

    limsup|an|1/n|x|2n/n = limsup|an|1/n|x|2 = |x|2/R < 1 So that the series converges

    Therefore, |x|2 < R [itex]\Rightarrow[/itex] |x| < [itex]\sqrt{R}[/itex]

    So R2 = [itex]\sqrt{R}[/itex] (Where R2 is the r.o.c. of the second series)


    So I tried doing this with the first series, but it doesn't work.

    So then I showed that limsup|bn|1/n2 < 1 implies that [itex]\sum[/itex]bn from n=0 to n=∞ converges absolutely (basically a modification of the root test)

    So then if I use this for the first series:

    limsup|an|1/n2|x|n2/n2 = limsup|an|1/n2|x| < 1 So that it converges.

    Therefore |x| < 1/limsup|an|1/n2 = R1

    But I'm assuming they want it in terms of R, so this doesn't really work...

    I feel like I'm close I'm just missing something?

    Any help would be greatly appreciated.
     
    Last edited: Apr 4, 2012
  2. jcsd
  3. Apr 3, 2012 #2

    Dick

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    I'm a little bit dubious about some this, and I think you meant your initial series to be a_n*x^n, not a_n*x^(n^2), but let's assume all of your limits exist. In the x^(2n) case you got |x|^2<R. In the x^(n^2) case you will get limit n->infinity |x|^n<R. What does that tell you about radius of convergence?
     
  4. Apr 3, 2012 #3
    How can we have |x|^n <R if n->infinity?

    Wouldn't it be 1/R*limsup|x|^n <1 ?

    So limsup|x|^n < R and wouldn't that only be true for |x| < 1 ?
     
  5. Apr 4, 2012 #4

    Dick

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    Yes, I think you need |x|<1.
     
  6. Apr 4, 2012 #5
    Oh ok, it just doesn't seem like a very formal way of showing it. Is there not a better way of doing it?
     
  7. Apr 4, 2012 #6

    Dick

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    Pick a number b such that |b|<1 and apply the root test to [itex]a_{n} b^{n^2}[/itex]. Now do the same for |b|>1.
     
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