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?
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.