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Power serie sum

  1. Jan 31, 2005 #1
    sum(x^n/(n²+2n),n=1..infinity)

    we've never heard of hypergeom functions and we encountered this on our exam. The question was not only the radius of convergence but the infinite sum as well. :surprised

    Anyone an idea?
     
  2. jcsd
  3. Jan 31, 2005 #2

    dextercioby

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    1.Factor the denominator:

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

    Write your series like:
    [tex] S(x)=\frac{1}{2}(\sum_{n=1}^{+\infty} \frac{x^{n}}{n}-\sum_{n=1}^{+\infty} \frac{x^{n}}{n+2}) [/tex]

    Daniel.

    P.S.I don't know how u can connect your series with the hypergeomtric ones.I'll think about it...
     
    Last edited: Jan 31, 2005
  4. Jan 31, 2005 #3

    arildno

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    Continue by writing:
    [tex]S(x)=\frac{1}{2}(\int_{0}^{x}\sum_{n=1}^{\infty}t^{n-1})dt-\frac{1}{x^{2}}\int_{0}^{x}(\sum_{n=1}^{\infty}t^{n+1})dt))=[/tex]
    [tex]\frac{1}{2}(\int_{0}^{x}\frac{dt}{1-t}-\frac{1}{x^{2}}\int_{0}^{x}\frac{t^{2}dt}{1-t})[/tex]
    or something like that..
     
    Last edited: Jan 31, 2005
  5. Jan 31, 2005 #4

    NateTG

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    You're moving terms in a conditionally convergent sequence which is not always kosher. Specifically, in this case, if [itex]x=1[/itex] the original series is convergent, while yours is not.

    I can suggest an alternative approach:
    As you pointed out:
    [tex] \frac{1}{n^{2}+2n}=\frac{1}{2}(\frac{x^n}{n}-\frac{x^n}{n+2}}) [/tex]
    This looks suspicously like a telescoping sum. Let's take a look at partial sums
    [tex]S(x,k)=\frac{1}{2}\sum_{n=1}^{k}\left(\frac{x^n}{n}-\frac{x^n}{n+2}}\right)[/tex]
    In the [itex]x=1[/itex] this is very nice:
    [tex]S(1,k)=\frac{1}{2}\left(\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+...\left(\frac{1}{k}-\frac{1}{k+2}\right)\right)[/tex]
    Now, the negative
    [itex]\frac{1}{3}[/itex]
    from the first term will cancel with the positve
    [itex]\frac{1}{3}[/itex]
    from the 3rd term. Similarly, for the negative elements in each of the following terms except for the last two. This means that the sum telescopes to:
    [tex]S(1,k)=\frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{k+1}-\frac{1}{k+2}\right)[/tex]
    so
    [tex]S(1)=\frac{3}{4}[/tex]

    It's pretty obvious that for [itex]|x|<1[/itex] the series are absolutely convergent (so Dex's approach will work) and for [tex]|x|>1[/tex] the series will be divergent since the individual terms will grow without bound.
     
    Last edited: Jan 31, 2005
  6. Jan 31, 2005 #5

    dextercioby

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    There's a typo
    [tex] S(1)=\frac{3}{4} [/tex]

    What do you mean by
    [tex] |x|<0 [/tex]

    Isn't the modulus ALWAYS REAL AND NONEGATIVE??

    Daniel.
     
    Last edited: Jan 31, 2005
  7. Jan 31, 2005 #6

    NateTG

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    Right, forgot the [itex]\frac{1}{2}[/itex].
    [/QUOTE]
    What do you mean by
    [tex] |x|<0 [/tex]
    Isn't the modulus ALWAYS REAL AND NONEGATIVE??[/QUOTE]

    I meant to say [itex]|x|<1[/itex] and [itex]|x|>1[/itex].


    (Will edit the post.)
     
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