TITLE: Proving a Lambert series identity from Ramanujan's Collected Papers
QUESTION [6 upvotes]: While studying Ramanujan's Collected Papers I came across an identity $$q(1 + q + q^{3} + q^{6} + \cdots)^{8} = \frac{1^{3}q}{1 - q^{2}} + \frac{2^{3}q^{2}}{1 - q^{4}} + \frac{3^{3}q^{3}}{1 - q^{6}} + \cdots$$ which I am unable to establish (although the identity looks simple). I tried to use the the following identities:
$\displaystyle Q(q) = 1 + 240\left(\frac{1^{3}q}{1 - q} + \frac{2^{3}q^{2}}{1 - q^{2}} + \frac{3^{3}q^{3}}{1 - q^{3}} + \cdots\right) = \left(\frac{2K}{\pi}\right)^{4}(1 + 14k^{2} + k^{4})$
$\displaystyle Q(-q) = \left(\frac{2K}{\pi}\right)^{4}(1 - 16k^{2} + 16k^{4})$
$\displaystyle Q(q^{2}) = \left(\frac{2K}{\pi}\right)^{4}(1 - k^{2} + k^{4})$
where we have $q = e^{-\pi K(k')/K(k)}$ and was trying to put the RHS (of the identity to be established) as a combination of the above $Q$ functions but was not able to do so. In this way I thought to transform the RHS in terms of $K, k$ and then reach the LHS which is $q\psi^{8}(q)$ where Ramanujan's $\psi(q)$ is related to Jacobi's theta function via $\vartheta_{2}(q) = 2q^{1/4}\psi(q^{2})$. I was able to write the RHS as
$\displaystyle \sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{2n}} = \frac{1}{2}\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{n}} + \frac{1}{2}\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 + q^{n}} = S_{1} + S_{2}$
where $S_{1}$ can be expresssed in terms of $Q(q)$, but somehow could not express $S_{2}$ in terms of $Q$ functions. Maybe I am on the wrong path. If there is any simpler way to prove the identity please let me know.

REPLY [6 votes]: 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.)