( Sum to infinity of a divisor function ) ^{2} simplify expression

Hi,

I have $120 \sum \limits_1^\infty (\sigma_{3}(n))^{2}$, where $\sigma_{3}(n)$ is the 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.

Can't necessarily help, but I'm struggling to make sense of that.

I presume $\displaystyle\sigma_3(n)=\sum_{d|n}d^3$

The sum you're given rapidly goes off to infinity, and you want to show it's equal to a a finite sum that is a function of n.

Agree with that, though, could it not be possible that n can go to infinity and this has just simplified an expression for each term?

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

From a bit of digging around here I came across:

$\sigma_7(n)-\sigma_3(n)= 120\sum \limits_{k=1}^{n-1} \sigma_{3}(k) \sigma_{3}(n-k)$
.

Yes, this is exactly what I am trying to show, since my question doesn't seem to make sense I will post the whole question instead to clarify things:

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

...

Yep, equating coeff. of $q^n$ gives:

$480\sigma_3(n)+(240)^2\sum \limits_{k=1}^{n-1}\sigma_3(k)\sigma_3(n-k) = 480\sigma_7(n)$

and divide by 480.

okay cheers, I see that.

However I'm struggling to write this neatly, so I've wrote out a bunch of terms and I can see the pattern, but my working is very improper, by any chance do you have any hints how to approach in a neat, proper manner?

thanks .