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A (sum to infinity of divisor function)^{2} -- simplify the expression

  1. Nov 7, 2016 #1
    Hi,



    I have ## 120 \sum \limits_1^\infty (\sigma_{3}(n))^{2} ## , where ## \sigma_{3}(n) ## is a divisor function.



    And I want to show that this can be written as ##120\sum \limits_{k=1}^{n-1} \sigma_{3}(k) \sigma_{3}(n-k) ##



    I'm pretty stuck on ideas starting of to be honest, since the sum is infinite, any help much appreciated.



    Many thanks in advance.
     
  2. jcsd
  3. Nov 8, 2016 #2

    stevendaryl

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    I'm wondering if there is a mistake in what you wrote. The first summation doesn't say what the index is. [itex]n[/itex]? There aren't any other indices to sum over.
     
  4. Nov 8, 2016 #3
    Okay so in that case I'm assuming I have made a mistake so I will give the whole question instead

    I have concluded that ##E_{4}(t){2}=E_{8}(t)## and I am wanting to use this to show that ##\sigma_{7}(n)=\sigma_{3}(n)+120\sum\limits^{n-1}_{k=1} \sigma_{3}(k)\sigma_{3}(n-k)##

    where ##E_{8}(t)=1+480\sum\limits^{\infty}_{n=1} \sigma_{7}(n)q^{n}## and ##E_{4}(t)=1+240\sum\limits^{\infty}_{n=1} \sigma_{3}(n)q^{n}##

    And so my working so far is

    ##E_{4}^{2}(t)-E_{8}(t)=0##
    ##(1+480\sum\limits^{\infty}_{n=1} \sigma_{7}(n)q^{n} -((1+480\sum\limits^{\infty}_{n=1} \sigma_{3}(n)q^{n})(1+480\sum\limits^{\infty}_{n=1} \sigma_{3}(n)q^{n}))

    = 480\sum\limits^{\infty}_{n=1} \sigma_{7}(n)q^{n} - 480\sum\limits^{\infty}_{n=1} \sigma_{3}(n)q^{n}-240^{2}\sum\limits^{\infty}_{k=1} \sigma_{3}(k)q^{k} \sum\limits^{\infty}_{n=1} \sigma_{3}(n)q^{n}##

    Divide by 480 and everything looks on track except this issue in my OP

    ##\sigma_{7}(n)q^{n}=q^{n}\sigma_{3}(n)+120\sum\limits^{\infty}_{k=1} \sigma_{3}(k)q^{k} \sum\limits^{\infty}_{n=1} \sigma_{3}(n)q^{n} ##
     
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