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

  • Thread starter binbagsss
  • Start date
1,155
6
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.
 

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.
 
1,155
6
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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