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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?

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It's actually a good excuse for you to learn a bit of modular forms. Modular forms are functions that behave well with respect to certain Mobius transformations. Moreover, the dimension of modular forms can be computed in some cases. In this case, the series on the left (without the square) is essentially the only modular form of weight 4. The series on the right is essentially the only modular form of weight 8. When you multiply two modular forms of weight 4, you get a modular form of weight 4+4 = 8. This forces LHS and RHS to be equal. – user27126 Jan 12 '13 at 2:18
I don't understand either the wikipedia or wolframalpha articles on modular forms. I don't really have any formal knowledge in the field of complex analysis either. – Ethan Jan 12 '13 at 2:24
If you are serious in math it's inevitable for you to have a much better general foundation. Perhaps the desire to understand identities like this would motivate you through the process. – user27126 Jan 12 '13 at 2:29
What do you mean by a 'general foundation'? – Ethan Jan 12 '13 at 2:30
Real/Complex/Functional Analysis, Linear/Abstract Algebra, Differential/Algebraic Topology, Differential/Algebraic Geometry etc, to name a few. It's not necessary for you to know all these right away, as this will take you a good few years. However, without sitting down and going through some basics of modern math it would be really difficult for you to keep going. – user27126 Jan 12 '13 at 2:33

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.

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This was the exact statement I was trying to prove with a proof of the previous statement lol. – Ethan Jan 13 '13 at 7:05

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.

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I am interested in this sort of thing, but I feel I need to study more before I try to learn more about elliptic functions and such, recently I have just been toying around with identitiess of Liouville, and using Skoruppa's identity found some interesting things, unfortunately I don't plan on studying this sort of thing for a while, I have only really been studying mathematics for about $2$ or $3$ years, and am 'lacking' in many other subjects which I would like to learn about. Though I saved a link to your blog for future reference, and I am looking forward to learning more about this. – Ethan Jul 11 '13 at 4:57

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