Lambert series help How can one go about proving lambert series identities like,
$$\left(1+240\sum_{n=1}^\infty \frac{n^3q^n}{1-q^n} \right)^2=1+480\sum_{n=1}^\infty \frac{n^7q^n}{1-q^n}$$
All the papers I have looked at require the knowledge of modular forms and other abstract mathematics, is it possible to prove an identity like this using only algebra? If not, could someone explain to me in laymen's terms, why the above identity holds?
 A: Ramanujan proved many such Lambert series identities using a very simple approach based on trigomometric identities. Unfortunately his work is shadowed by the modern techniques of complex analysis and modular forms. Please refer his paper "On Certain arithmetical functions" which appeared in Transactions of the Cambridge Philosophical Society in 1916. I have written blog posts on the same. This particular identity under discussion is one of the simplest.
A: It turns out that many of these Lambert series identities can be proved without modular forms using the Huard/Ou/Spearman/Williams theorem, which is here. First expand the right side to obtain
$$ 1 + 480 \sum_{n\ge 1} \frac{n^7 q^n}{1-q^n} = 
1 + 480 \sum_{m\ge 1} n^7 \sum_{k\ge 1} q^{kn} =
1 + 480 \sum_{m\ge 1} \left( \sum_{d|m} d^7 \right) q^m $$
which is $$1 + 480 \sum_{m\ge 1} \sigma_7(m) q^m.$$
By the same reasoning the left side is
$$ \left( 1 + 240 \sum_{m\ge 1} \sigma_3(m) q^m \right)^2
= 1 + 480 \sum_{m\ge 1} \sigma_3(m) q^m +
240^2 \sum_{m\ge 1} q^m \sum_{k=1}^{m-1} \sigma_3(k) \sigma_3(m-k).$$
So what we need to show here is
$$ 480 \sigma_3(m) + 240^2 \sum_{k=1}^{m-1} \sigma_3(k) \sigma_3(m-k) =
480 \sigma_7(m) $$ or
$$\sigma_7(m) = \sigma_3(m) + 120 \sum_{k=1}^{m-1} \sigma_3(k) \sigma_3(m-k).$$
This is precisely identity 3.17 from the above paper, where a proof is given.
