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

#### binbagsss

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.

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#### stevendaryl

Staff Emeritus
I'm wondering if there is a mistake in what you wrote. The first summation doesn't say what the index is. $n$? There aren't any other indices to sum over.

#### binbagsss

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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