Proving a Lambert series identity from Ramanujan's Collected Papers

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.

• Using $Q(q)$ was more to the point. $R(q)$ uses $n^5$, not $n^3$. – ccorn Jul 9 '13 at 11:33
• Sorry ccorn, it seems I got carried away due to some other thoughts. Now it is back to the earlier version. Hopefully it will not cause any further confusion. By the way I liked the rollback feature in editing. – Paramanand Singh Jul 9 '13 at 11:36
• It might already help you if you use not the "plus" partial fraction decomposition but the "minus" one which delivers a factor $q^n$ for free in the expansion of $Q(q) - Q(q^2)$. – ccorn Jul 9 '13 at 11:51
• ccorn, but in that case the individual series would not converge as $n^{3}/(1 - q^{n})$ wont tend to zero as $n \to \infty$. I had first thought of that but then rejected it. – Paramanand Singh Jul 9 '13 at 11:54
• Yes, that's why I added "in $Q(q)-Q(q^2)$" afterwards, sorry. – ccorn Jul 9 '13 at 11:55

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.
• Note: When thinking in terms of functions instead of formal power series, we have to assume that $|q|<1$ so the series converge absolutely. – ccorn Jul 9 '13 at 13:03