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Ramanujan presented many identities, Hardy chose one which for him represented the best of Ramanujan. There are many proofs for this identity.

(for example, H. H. Chan’s proof, M. Hirschhorn's proof...)

Is there an elementary proof for Ramanujan's "most beautiful" identity?

$$\displaystyle{\sum_{n=0}^\infty p(5n+4)q^n}=5\frac{(q^5;q^5)^5_\infty}{(q;q)^6_\infty}$$ for $|q|<1$, where $p(n)$ is the partition function.

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    $\begingroup$ See exercises 14.11 through 14.15 here. They start towards the bottom of page 180. $\endgroup$ – Chip Hurst Sep 15 '16 at 14:28
  • $\begingroup$ @ChipHurst: thanks for the link ... $\endgroup$ – user354674 Oct 5 '16 at 19:23
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    $\begingroup$ @mike4ty4: The beauty is in the surprising nature of the equation. That there should be an infinite product expression for the series $\sum p(5n+4)q^{n}$ is not obvious. And such formulas are rare. There is only one more similar formula for $\sum p(7n+5)q^{n}$. See details in my answer. $\endgroup$ – Paramanand Singh Jan 31 '17 at 5:24
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The original proof by Ramanujan is available in my blog post (see equation $(17)$). An interesting and much simpler proof is provided by Oddmund Kolberg and I have presented his technique for the more complicated identity $$\begin{align}\sum_{n = 0}^{\infty}p(7n + 5)q^{n}&= 7\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{3}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{4}}\notag\\ &\,\,\,\,\,\,\,\,+ 49q\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{7}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{8}}\tag{1}\end{align}$$ in another blog post.

The same technique can be used to prove the identity in question here and with considerably less effort. Let $$\phi(q) = \prod_{n = 1}^{\infty}(1 - q^{n}) = (q; q)_{\infty}\tag{2}$$ We will make use of the identities \begin{align} \phi(q) &= \sum_{n = -\infty}^{\infty}(-1)^{n}q^{n(3n + 1)/2}\tag{3}\\ \phi^{3}(q) &= \sum_{n = 0}^{\infty}(-1)^{n}(2n + 1)q^{n(n + 1)/2}\tag{4}\\ \frac{1}{\phi(q)} &= \sum_{n = 0}^{\infty}p(n)q^{n}\tag{5} \end{align} Next step is to split the sums in $(3), (4), (5)$ based on powers of $q$ modulo $5$ so that $$\phi(q) = g_{0} + g_{1} + g_{2} + g_{3} + g_{4}\tag{6}$$ where $$g_{s} = \sum_{n = -\infty, n(3n + 1)/2 \equiv s \pmod{5}}^{\infty}(-1)^{n}q^{n(3n + 1)/2}\tag{7}$$ and $$\phi^{3}(q) = h_{0} + h_{1} + h_{2} + h_{3} + h_{4}\tag{8}$$ where $$h_{s} = \sum_{n = 0,n(n + 1)/2\equiv s\pmod{5}}^{\infty}(-1)^{n}(2n + 1)q^{n(n + 1)/2}\tag{9}$$ and $$\frac{1}{\phi(q)} = P_{0} + P_{1} + P_{2} + P_{3} + P_{4}\tag{10}$$ where $$P_{s} = \sum_{n = 0}^{\infty}p(5n + s)q^{5n + s}\tag{11}$$ Multiplying equations $(6)$ and $(10)$ and combining terms based on powers of $q$ modulo $5$ we get the following set of equations \begin{align} g_{0}P_{0} + g_{4}P_{1} + g_{3}P_{2} + g_{2}P_{3} + g_{1}P_{4} &= 1\notag\\ g_{1}P_{0} + g_{0}P_{1} + g_{4}P_{2} + g_{3}P_{3} + g_{2}P_{4} &= 0\notag\\ g_{2}P_{0} + g_{1}P_{1} + g_{0}P_{2} + g_{4}P_{3} + g_{3}P_{4} &= 0\notag\\ g_{3}P_{0} + g_{2}P_{1} + g_{1}P_{2} + g_{0}P_{3} + g_{4}P_{4} &= 0\notag\\ g_{4}P_{0} + g_{3}P_{1} + g_{2}P_{2} + g_{1}P_{3} + g_{0}P_{4} &= 0\notag \end{align} Our goal to is calculate $P_{4}$ which is given by $P_{4} = D_{4}/D$ where $D_{4}, D$ are determinants given by $$D = \begin{vmatrix}g_{0} & g_{4} & g_{3} & g_{2} & g_{1}\\ g_{1} & g_{0} & g_{4} & g_{3} & g_{2}\\ g_{2} & g_{1} & g_{0} & g_{4} & g_{3}\\ g_{3} & g_{2} & g_{1} & g_{0} & g_{4}\\ g_{4} & g_{3} & g_{2} & g_{1} & g_{0}\end{vmatrix}\tag{12}$$ and $$D_{4} = \begin{vmatrix}g_{1} & g_{0} & g_{4} & g_{3}\\ g_{2} & g_{1} & g_{0} & g_{4}\\ g_{3} & g_{2} & g_{1} & g_{0}\\ g_{4} & g_{3} & g_{2} & g_{1}\end{vmatrix}\tag{13}$$ The above determinants are evaluated using some relations between the $g_{s}$'s. First of all note that by definition of $g_{s}$ we have $g_{3} = g_{4} = 0$ (because $n(3n + 1)/2$ is never equal to $3$ or $4$ modulo $5$) and since $n(3n + 1)/2 \equiv 1 \pmod {5}$ is equivalent to $n \equiv -1 \pmod{5}$ we have \begin{align} g_{1} &= \sum_{n = -\infty}^{\infty}(-1)^{5n - 1}q^{(5n - 1)(15n - 2)/2}\notag\\ &= -\sum_{n = -\infty}^{\infty}(-1)^{n}q^{1 + 25n(3n - 1)/2}\notag\\ &= -q\sum_{n = -\infty}^{\infty}(-1)^{n}q^{25n(3n + 1)/2}\notag\\ &= -q\phi(q^{25})\tag{14} \end{align} Similarly $n(n + 1)/2$ is never equal to $2$ modulo $5$ and hence $h_{2} = 0$. From $(3)$ and $(4)$ we have $$\sum h_{s} = \left(\sum g_{s}\right)^{3}$$ or $$\sum h_{s} = (g_{0} + g_{1} + g_{2})^{3}$$ and equating the terms with powers of $q$ equal to $2$ modulo $5$ we get $$h_{2} = 3g_{0}^{2}g_{2} + 3g_{1}^{2}g_{0}$$ Noting that $h_{2} = 0$ we get $$g_{0}g_{2} + g_{1}^{2} = 0\tag{15}$$ The evaluation of the determinant $D_{4}$ is considerably simplified if we use the notation $$\alpha = g_{0}/g_{1}, \beta = g_{2}/g_{1}$$ From $(15)$ we get $$\alpha\beta = -1\tag{16}$$ and the determinant $D_{4}$ gets simplified as $$D_{4} = g_{1}^{4}\begin{vmatrix}1 & \alpha & 0 & 0\\ \beta & 1 & \alpha & 0\\ 0 & \beta & 1 & \alpha\\ 0 & 0 & \beta & 1\end{vmatrix} = 5g_{1}^{4}\tag{17}$$ The evaluation of determinant $D$ is aided by the fact that it is the determinant of a circulant matrix $A$. The determinant of a square matrix is the product of its eigenvalues and it is easy to find the eigenvalues of a circulant matrix. If $\omega$ is a $5^{\text{th}}$ root of unity (including $1$) then $$g_{0} + \omega g_{1} + \omega^{2}g_{2} + \omega^{3}g_{3} + \omega^{4}g_{4}$$ is an eigenvalue of $A$. Thus if $\omega$ is a primitive $5^{\text{th}}$ root of unity then $$\lambda_{t} = g_{0} + \omega^{t} g_{1} + \omega^{2t}g_{2} + \omega^{3t}g_{3} + \omega^{4t}g_{4}$$ gives all the eigenvalues of $A$ for $t = 0, 1, 2, 3, 4$. The determinant $D$ is therefore given by $$D = \prod_{t = 0}^{4}(g_{0} + \omega^{t} g_{1} + \omega^{2t}g_{2} + \omega^{3t}g_{3} + \omega^{4t}g_{4} = \prod_{t = 0}^{4}\sum_{s = 0}^{4}\omega^{st}g_{s}$$ From the definition of $g_{s} = g_{s}(q)$ we can easily see that $\omega^{st}g_{s}(q) = g_{s}(\omega^{t}q)$ and hence \begin{align} D &= \prod_{t = 0}^{4}\sum_{s = 0}^{4}\omega^{st}g_{s} = \prod_{t = 0}^{4}\sum_{s = 0}^{4}g_{s}(\omega^{t}q) = \prod_{t = 0}^{4}\phi(\omega^{t}q)\notag\\ &= \prod_{t = 0}^{4}\prod_{n = 1}^{\infty}(1 - \omega^{nt}q^{n}) = \prod_{n = 1}^{\infty}\prod_{t = 0}^{4}(1 - \omega^{tn}q^{n})\notag\\ &= \prod_{n \not\equiv 0\pmod{5}}(1 - q^{5n})\prod_{n \equiv 0\pmod{5}}(1 - q^{n})^{5} = \frac{\phi^{6}(q^{5})}{\phi(q^{25})}\tag{18} \end{align} From equations $(14), (17)$ and $(18)$ we can see that $$\sum_{n = 0}^{\infty}p(5n + 4)q^{5n + 4} = P_{4} = \frac{D_{4}}{D} = 5q^{4}\frac{\phi^{5}(q^{25})}{\phi^{6}(q^{5})}$$ and replacing $q^{5}$ by $q$ we get $$\sum_{n = 0}^{\infty}p(5n + 4)q^{n} = 5\frac{\phi^{5}(q^{5})}{\phi^{6}(q)}=5\frac{(q^{5};q^{5})_{\infty}^{5}}{(q;q)_{\infty}^{6}}\tag{19}$$ which is the desired result in question.

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