Ramanujan's Generating Function for Partitions Modulo 7

Ramanujan's Partition Congruences

Based on the empirical analysis of a table of partitions Ramanujan conjectured his famous partition congruences $$\boxed{\begin{align}p(5n + 4)&\equiv 0\pmod{5}\notag\\ p(7n + 5)&\equiv 0\pmod{7}\notag\\ p(11n + 6)&\equiv 0\pmod{11}\notag\end{align}}\tag{1}$$ and gave some of the most beautiful proofs for them (see here). In addition to these proofs he gave the following generating functions for $p(5n + 4), p(7n + 5)$: $$\sum_{n = 0}^{\infty}p(5n + 4)q^{n} = 5\frac{\{(1 - q^{5})(1 - q^{10})(1 - q^{15})\cdots\}^{5}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{6}}\tag{2}$$ and \begin{align}\sum_{n = 0}^{\infty}p(7n + 5)q^{n}&= 7\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{3}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{4}}\notag\\ &\,\,\,\,\,\,\,\,+ 49q\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{7}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{8}}\tag{3}\end{align} We have already established $(2)$ in one of our posts and this post deals with the identity $(3)$ concerning generating function of partitions modulo $7$.

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A Continued Fraction for Error Function by Ramanujan

Today's post is inspired by this question I asked sometime back on MSE. And after some effort (offering a bounty) I received a very good answer from a user on MSE. This question was asked for the first time by Ramanujan in the "Journal of Indian Mathematical Society" 6th issue as Question no. 541, page 79 in the following manner:

Prove that $$\left(1 + \frac{1}{1\cdot 3} + \frac{1}{1\cdot 3\cdot 5} + \cdots\right) + \left(\cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}\right) = \sqrt{\frac{\pi e}{2}}\tag{1}$$ It turns out that the first series is intimately connected with the error function given $$\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2}\,dt\tag{2}$$ In his famous letter to G. H. Hardy, dated 16th January 1913, Ramanujan gave the following continued fraction for the integral used in the definition of error function given above: $$\int_{0}^{a}e^{-x^{2}}\,dx = \frac{\sqrt{\pi}}{2} - \cfrac{e^{-a^{2}}}{2a+}\cfrac{1}{a+}\cfrac{2}{2a+}\cfrac{3}{a+}\cfrac{4}{2a+\cdots}\tag{3}$$

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Playing With Partitions: Euler's Pentagonal Theorem

This post was originally written for MSE blog. Please click here to read.

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