Sometime back I asked question regarding a Lambert series identity from Ramanujan's Collected Papers and with hints provided I was able to prove it easily. Now I have encountered another identity (in Collected Papers) of similar nature namely:

$\displaystyle \frac{1^{5}q}{1 - q} + \frac{3^{5}q^{2}}{1 - q^{3}} + \frac{5^{5}q^{3}}{1 - q^{5}} + \cdots = \left\{1 + 240\left(\frac{1^{3}q}{1 - q} + \frac{2^{3}q^{2}}{1 - q^{2}} + \cdots\right)\right\}\left(\frac{q}{1 - q} + \frac{3q^{2}}{1 - q^{3}} + \frac{5q^{3}}{1 - q^{5}} + \cdots\right)$

and this one seems more difficult to handle as there are odd exponents in denominator due to which splitting into partial fraction does not seem possible. Any hints or a solution would be highly appreciated.

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    $\begingroup$ Using the function defs from the previous question and setting $q=q_2^2$, this is the identity $\frac{q_2}{1008}\left(R(-q_2) - R(q_2)\right) = Q(q)\,q\,\psi^4(q)$. I suppose you can work this out now. Unless you or someone else posts an answer, I'll work out a brief demonstration of the identity. $\endgroup$ – ccorn Jul 11 '13 at 10:07
  • $\begingroup$ Oops, Ramanujan's $R(q)$ has not been introduced yet: $R(q) = 1 - 504\sum_{n=1}^\infty \frac{n^5\,q^n}{1-q^n}$ $\endgroup$ – ccorn Jul 11 '13 at 10:17
  • $\begingroup$ Thanks ccorn for the hint. I can proceed without any problem. $\endgroup$ – Paramanand Singh Jul 11 '13 at 13:05

Based on hint from ccorn (see comments) here is the solution to the problem asked. First from equation $(7)$ of this post we can see that $$\psi^{4}(q) = \sum_{n = 0}^{\infty}\frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}$$ and therefore the RHS of the identity to be established can be written as $Q(q)q\psi^{4}(q)$.

The LHS on the other hand looks complicated but using definition of $R(q)$ we can see that $$R(-q) - R(q) = 1008\sum_{n = 0}^{\infty}\frac{(2n + 1)^{5}q^{2n + 1}}{1 - q^{4n + 2}} = \frac{1008}{q}\sum_{n = 0}^{\infty}\frac{(2n + 1)^{5}(q^{2})^{n + 1}}{1 - (q^{2})^{2n + 1}}$$ so that the LHS is easily seen to be $$\frac{\sqrt{q}\{R(-\sqrt{q}) - R(\sqrt{q})\}}{1008}$$ To evaluate this expression it is better to evaluate $R(-q) - R(q)$ first and replace $q$ by $\sqrt{q}$ later. We know that (from here) $$R(-q) = \left(\frac{2K}{\pi}\right)^{6}(1 - 2k^{2})(1 + 32k^{2} - 32k^{4}),\, R(q) = \left(\frac{2K}{\pi}\right)^{6}(1 + k^{2})(1 - 34k^{2} + k^{4})$$ so that we have $$R(-q) - R(q) = 63k^{2}\left(\frac{2K}{\pi}\right)^{6}(1 - k^{2} + k^{4})$$ Replacing $q$ by $\sqrt{q}$ in above (thereby replacing $K$ by $(1 + k)K$ and $k$ by $2\sqrt{k}/(1 + k)$) we get \begin{align} R(-\sqrt{q}) - R(\sqrt{q}) &= 252k\left(\frac{2K}{\pi}\right)^{6}(1 + 14k^{2} + k^{4})\notag\\ &= 252k\left(\frac{2K}{\pi}\right)^{2}Q(q)\notag\\ &= 252\theta_{2}^{2}(q)\theta_{3}^{2}(q)Q(q)\notag\\ &= 1008\sqrt{q}\psi^{2}(q^{2})\phi^{2}(q)Q(q)\notag\\ &= 1008\sqrt{q}\psi^{4}(q)Q(q)\notag \end{align} so that we have $$\frac{\sqrt{q}\{R(-\sqrt{q}) - R(\sqrt{q})\}}{1008} = Q(q)q\psi^{4}(q)$$ which is what we had to prove.

Normally I try to prove such identities concerning Lambert series by converting them to formulas containing $K, k$ and then doing algebraic manipulations. I wish there were other methods by which one could prove identities by direct manipulation of the Lambert series without going through the elliptic integral $K$.

Ramanujan just wrote that such identities can be easily established either by elliptic function theory or by elementary methods. One possible approach is to prove the corresponding identity concerning divisor sum functions obtained by equating coefficients of $q^{n}$ on both sides but this itself requires some number theory. What we need is some proof just based on algebraic manipulation of the Lambert series.

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    $\begingroup$ Instead of using $K$ and $k$, one can use expressions in terms of Jacobi's thetanull functions which yield more symmetric expressions. Moreover, symmetric polynomials of all three Jacobi thetanull functions have some modular symmetries which allow proofs like: "After suitable rearrangement, both sides can easily be shown to be holomorphic modular forms of weight 4 (say); knowing that the dimension of such function space is 1 (say), we only need to check that the first 1 series term(s) of both sides agree; this is the case." My collection of identities thus obtained is still growing. $\endgroup$ – ccorn Jul 11 '13 at 21:13
  • $\begingroup$ Now demonstrated for a related question. Such proofs take space the very first time, but they ease things a lot thereafter. $\endgroup$ – ccorn Jul 17 '13 at 23:37

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