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Hi everyone, How I can calculated this.. $$\left(1-24\sum_{n=1}^{\infty}{\dfrac{n}{1-x^n}x^n}\right)\left(1+240\sum_{n=1}^{\infty}{\dfrac{n^3}{1-x^n}x^n}\right)$$

My greatest problem is how find $$\left(\sum_{n=1}^{\infty}{\dfrac{n}{1-x^n}x^n}\right)\left(\sum_{n=1}^{\infty}{\dfrac{n^3}{1-x^n}x^n}\right)$$ Yes, I know that $$\left(\sum_{n=1}^{\infty}{a_nx^n}\right)\left(\sum_{n=1}^{\infty}{b_nx^n}\right)=\left(\sum_{n=1}^{\infty}{c_nx^n}\right)$$ Where $c_n=\sum_{i=1}^{\infty}{a_nb_{n-i}x^n}$ Can give some hint, thanks!

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    Your original series are not formal power series. If you product also need not be a power series, you can just use $$ a_{n} = {n \over 1 - x^{n}}, \quad b_{n} = {n^3 \over 1 - x^{n}}, $$ without necessarily trying to collect all the powers of $x$. – 2017-02-08
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    Have a look at Ramanujan's identities for Eisenstein series: https://en.wikipedia.org/wiki/Eisenstein_series – 2017-02-08

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By setting $x=q$ and using Ramanujan's notation, you are asking for $LM$.
Since $LM=3q\frac{dM}{dq}+N$ (by the theory of modular forms for $\text{SL}(2,\mathbb{Z})$) $$ LM = 1+720\sum_{n\geq 1}n\sigma_3(n)q^{n}-504\sum_{n\geq 1}\sigma_{5}(n) q^n. $$