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Sum of power series

  1. Feb 23, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that the power series [itex]\sum_{k=1}^{k=\infty} \frac{x^{2k+1}}{k(2k+1)}[/itex] converges uniformly when [itex]|x| \leq 1[/itex]and determine the sum (at least when [itex]|x| < 1[/itex]).

    3. The attempt at a solution

    Couldn't I somehow go about and show that, as [itex]|x| \leq 1[/itex], then [itex]f = \frac{x^{2k+1}}{k(2k+1)} \leq \frac{1^{2k+1}}{k(2k+1)} = \frac{1}{k(2k+1)} = g < \frac{1}{2k^2}[/itex] which converges. Then by Weierstrass majorant principle f should converge. Or am I missing something? And how do I find this sum? :confused:
  2. jcsd
  3. Feb 23, 2007 #2
    you should use the ratio test. where ever the limit evaluated by ratio test is less than one the series converges

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  4. Feb 24, 2007 #3
    It seems from the way you write your explanation you confuse the sum with the individual terms in the sum.

    Anyway, missing some absolute values you have a handle on showing that terms are each bounded (in absolute value) uniformly in x as indicated.

    To see what the sum is, what you get if you differentiated term by term?
    You would have to manipulate factors and powers of x, but it reduces to using a Taylor series you know.
  5. Feb 27, 2007 #4
    But if I use the ratio test, is a absolutely convergent sum also uniformly convergent?
  6. Feb 28, 2007 #5

    That it converges is one thing. What sum of the infinite series turns out for x as indicated is another
  7. Feb 28, 2007 #6
    I'm sorry but you're speaking in riddles for me... :blushing: :uhh:
  8. Feb 28, 2007 #7
    These are excerpts from the question:

  9. Feb 28, 2007 #8
    Sorry, with "And how do I find this sum?" I was referring to the power series in section 1.
  10. Feb 28, 2007 #9
    As far as I understand it I can't use the ratio test to show uniform convergence.
  11. Feb 28, 2007 #10


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    You certainly should know that if a power series converges absolutely on a set, then it converges uniformly on that set. See\

    He was just pointing out that you have two questions- where the series converges uniformly and what the sum is. It might help to remember that any power series is the Taylor's series for the function it converges to.

    Here's what I would do. First calculate a few terms to see what is happening:
    [tex]\frac{x^3}{3}+ \frac{x^5}{10}+ \frac{x^7}{21}+ \frac{x^9}{36}+\cdot\cdot\cdot + \frac{x^{2k+1}}{k(2k+1)}+ \cdot\cdot\cdot[/tex]
    Factor out x3:
    [tex]= x^3\left(\frac{1}{3}+ \frac{x^2}{10}+ \frac{x^4}{21}+ \frac{x^6}{36}+\cdot\cdot\cdot + \frac{x^{2(k-1)}}{k(2k+1)}+ \cdot\cdot\cdot\right)[/tex]
    Write that as powers of x2:
    [tex]= x^3\left(\frac{1}{3}+ \frac{x^2}{10}+ \frac{(x^2)^2}{21}+ \frac{(x^2)^3}{36}+\cdot\cdot\cdot + \frac{(x^2)^{k-1}}{k(2k+1)}+ \cdot\cdot\cdot\right)[/tex]

    Can you think of a function that has Taylor's series
    [tex]\frac{1}{3}+ \frac{x}{10}+ \frac{x^2}{21}+ \frac{x^3}{36}+\cdot\cdot\cdot + \frac{x^{k-1}}{k(2k+1)}+ \cdot\cdot\cdot[/tex]?
  12. Mar 1, 2007 #11
    Ah... I totally missed that I had written "sum" instead of "series" in post #4. Guess this scheme of mine, trying to work full days and then brushing up on my math during the nights isn't going to well. I'm just too tired most of the time. My apologies.

    Not really...
  13. Mar 1, 2007 #12
    As in my previous installment, if you differentiated term by term (maybe more that once, if needed), you would arrive at a more familiar series. Then work backwards (integrate) so you get an explicit formula.
  14. Mar 2, 2007 #13
    heh, you want to try to find some close form for the original series??? just differentiate with respect to x!!! (it is quite obvious! come on!)
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