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Math Help - Showing two infinite sums are equal?

  1. #1
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    Showing two infinite sums are equal?

    I have this problem requiring me to show two infinite sums are equal but I can't seem to figure out how to do this. No matter what I try never give any results.

    Hence my question is what approach should I use toward one of these question?

    This is the question in question if it can help


    Show
    <br />
\sum_{p=0}^{\infty}\frac{x^{2p}}{1+x^{4p+2}}=\sum_  {q=0}^{\infty} (-1)^{q}\frac{x^{2q}}{1-x^{4q+2}}<br />

    if |x|<1
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  2. #2
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    Quote Originally Posted by KZA459 View Post
    I have this problem requiring me to show two infinite sums are equal but I can't seem to figure out how to do this. No matter what I try never give any results.

    Hence my question is what approach should I use toward one of these question?

    This is the question in question if it can help


    Show
    <br />
\sum_{p=0}^{\infty}\frac{x^{2p}}{1+x^{4p+2}}=\sum_  {q=0}^{\infty} (-1)^{q}\frac{x^{2q}}{1-x^{4q+2}}<br />

    if |x|<1
    Use the binomial series for (1+x^{4p+2})^{-1} to see that \sum_{p=0}^{\infty}\frac{x^{2p}}{1+x^{4p+2}} = \sum_{p=0}^{\infty}x^{2p}\sum_{q=0}^{\infty}(-1)^qx^{(4p+2)q} = \sum_{p=0}^{\infty}\sum_{q=0}^{\infty}(-1)^qx^{4pq+2p+2q}. Reverse the order of summation (why is this justified?) and then unwind the double sum to get \sum_{q=0}^{\infty} (-1)^{q}\frac{x^{2q}}{1-x^{4q+2}}.
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  3. #3
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    Darn, thanks a lot Opalg.
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  4. #4
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    Sorry, me again.

    I also have the show a similar result for |x|>1

    However using the power serie of \frac{1}{1+x^{4p+2}}
    does not work in this case since it has as radius of convergence 1.

    I must show if |x|>1 then
    <br />
\sum_{p=0}^{\infty}\frac{x^{2p}}{1+x^{4p+2}}= -\sum_{q=0}^{\infty} (-1)^{q}\frac{x^{2q}}{1-x^{4q+2}}<br />

    Unfortunately I am still as stuck and can't find a way to figure that one out even with the insight you gave me Olpag.

    Anyone has an idea?
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  5. #5
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    Quote Originally Posted by KZA459 View Post
    I also have the show a similar result for |x|>1

    However using the power serie of \frac{1}{1+x^{4p+2}}
    does not work in this case since it has as radius of convergence 1.

    I must show if |x|>1 then
    <br />
\sum_{p=0}^{\infty}\frac{x^{2p}}{1+x^{4p+2}}= -\sum_{q=0}^{\infty} (-1)^{q}\frac{x^{2q}}{1-x^{4q+2}}<br />

    Unfortunately I am still as stuck and can't find a way to figure that one out
    If you want a series that converges when |x| > 1 then you need it to use negative powers of x. So write \frac{x^{2p}}{1+x^{4p+2}}= x^{-2}\frac{(x^{-1})^{2p}}{1+(x^{-1})^{4p+2}} (dividing top and bottom by x^{4p+2}). You can then apply the previous result with x^{-1} in place of x, noticing that an extra minus sign will come in when you transform it back into positive powers of x.
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