Proving a Lambert series identity from Ramanujan's Collected Papers

Solution 1:

Paramanand Singh already found the answer. Nevertheless I'll try to post an answer here.

To avoid issues with branch cuts of $K$, let us replace $$\frac{2K}{\pi} = \vartheta_3^2(q);\quad k = \frac{\vartheta_2^2(q)}{\vartheta_3^2(q)}$$ where $$\vartheta_3(q) = \sum_{n\in\mathbb{Z}} q^{n^2}$$ is another Jacobi thetanull function. Thus $$Q(q)-Q(q^2) = 15\,\vartheta_2^4(q)\,\vartheta_3^4(q)$$

Now use the doubling formula: $$\vartheta_2^2(q) = 2\,\vartheta_2(q^2)\,\vartheta_3(q^2)$$

Therefore $$Q(q)-Q(q^2) = \frac{15}{16}\vartheta_2^8(q^{1/2}) = 240\,q\,\psi^8(q)$$

Consequently, your LHS equals $$q\,\psi^8(q) = \frac{1}{240}\left(Q(q)-Q(q^2)\right) =\sum_{n=1}^\infty\left(\frac{n^3\,q^n}{1-q^n} -\frac{n^3\,q^{2n}}{1-q^{2n}}\right)$$ $$= \sum_{n=1}^\infty n^3\,\frac{q^n\,(1+q^n)-q^{2n}}{1-q^{2n}} = \sum_{n=1}^\infty \frac{n^3\,q^n}{1-q^{2n}}$$ which is the RHS, QED.

(Edit: typo corrections.)