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Plouffe established the following formula for $\pi$ in $2006$ $$\pi = 72\sum_{n = 1}^{\infty}\frac{1}{n(e^{n\pi} - 1)} - 96\sum_{n = 1}^{\infty}\frac{1}{n(e^{2n\pi} - 1)} + 24\sum_{n = 1}^{\infty}\frac{1}{n(e^{4n\pi} - 1)}$$ I was trying to establish this by analyzing the function $$a(q) = \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})}$$ where $q = e^{-\pi}$. The formula given by Plouffe is equivalent to $$\pi = 72 a(q) - 96a(q^{2}) + 24a(q^{4})$$ I wonder if there is an expression for $a(q)$ in terms of $K, k$ where $k = \vartheta_{2}^{2}(q)/\vartheta_{3}^{2}(q)$ and $K = \pi\vartheta_{3}^{2}(q)/2$ and $q = e^{-\pi K'/K}$. Since $q = e^{-\pi}$ this would imply that $K' = K$ and hence $k = k' = 1/\sqrt{2}$.

Let me know if it is possible to express $a(q)$ in the desired form as above. Or any other proof of the Plouffe's formula would be greatly appreciated.

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  • $\begingroup$ Evidently, $$a(q)=\sum_{n=1}^\infty \frac{\sigma(n)}{n}q^n$$ Not sure if that helps. $\endgroup$ – Thomas Andrews Sep 19 '14 at 18:19
  • $\begingroup$ @ThomasAndrews: Actually I want $a(q)$ to be expressed in terms of $(q;q)_{\infty}$ $\endgroup$ – Paramanand Singh Sep 19 '14 at 18:20
  • $\begingroup$ @ThomasAndrews: finally found the answer myself and posted it. Its bit complicated but if one knows the theory of elliptic integrals and theta functions then the calculations are easily understood. $\endgroup$ – Paramanand Singh Sep 20 '14 at 5:14
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After a little more thinking I got the function $a(q)$ in the form I wanted. We start from the equation $$\eta(q) = q^{1/24}\prod_{n = 1}^{\infty}(1 - q^{n}) = 2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}$$ Taking logs we get $$\frac{\log q}{24} + \sum_{n = 1}^{\infty}\log(1 - q^{n}) = \frac{\log k}{12} + \frac{\log k'}{3} + \frac{\log 2}{3} + \frac{1}{2}\log\left(\frac{K}{\pi}\right)$$ or $$-\frac{\pi}{24}\cdot\frac{K'}{K} - \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})} = \frac{\log k}{12} + \frac{\log k'}{3} + \frac{\log 2}{3} + \frac{1}{2}\log\left(\frac{K}{\pi}\right)$$ so that the function $a(q)$ is given by $$a(q) = \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})} = -\frac{\log k}{12} - \frac{\log k'}{3} - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{24K}\tag{1}$$ Now if we change $q$ to $q^{2}$ we need to replace $k$ by $(1 - k')/(1 + k')$ and $K$ by $K(1 + k')/2$ and the ratio $K'/K$ just doubles up so we get \begin{align}a(q^{2}) &= -\frac{1}{12}\log\left(\frac{1 - k'}{1 + k'}\right) - \frac{1}{6}\log\left(\frac{4k'}{(1 + k')^{2}}\right) - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K(1 + k')}{2\pi}\right) - \frac{\pi K'}{12K}\notag\\ &= -\frac{1}{6}\log\left(\frac{k}{1 + k'}\right) - \frac{1}{6}\log\left(\frac{4k'}{(1 + k')^{2}}\right) - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K(1 + k')}{2\pi}\right) - \frac{\pi K'}{12K}\notag\\ &= -\frac{1}{6}\log\left(\frac{4kk'}{(1 + k')^{3}}\right) - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K(1 + k')}{2\pi}\right) - \frac{\pi K'}{12K}\notag\\ &= -\frac{\log(4kk')}{6} - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K}{2\pi}\right) - \frac{\pi K'}{12K}\notag\\ &= -\frac{\log(kk')}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{12K}\notag \end{align} To summarize $$a(q^{2}) = -\frac{\log(kk')}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{12K}\tag{2}$$ Again replacing $q$ by $q^{2}$ we get \begin{align} a(q^{4}) &= -\frac{1}{6}\log\left(\frac{1 - k'}{1 + k'}\right) - \frac{1}{12}\log\left(\frac{4k'}{(1 + k')^{2}}\right) - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K(1 + k')}{2\pi}\right) - \frac{\pi K'}{6K}\notag\\ &= -\frac{1}{3}\log\left(\frac{k}{1 + k'}\right) - \frac{1}{12}\log\left(\frac{4k'}{(1 + k')^{2}}\right) - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K(1 + k')}{2\pi}\right) - \frac{\pi K'}{6K}\notag\\ &= -\frac{1}{12}\log(4k^{4}k') - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{2\pi}\right) - \frac{\pi K'}{6K}\notag\\ &= -\frac{1}{12}\log(k^{4}k') + \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{6K}\notag \end{align} so that $$a(q^{4}) = -\frac{1}{12}\log(k^{4}k') + \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{6K}\tag{3}$$ Now time to do some algebra. Let us put $q = e^{-\pi}$ so that $k = k' = 2^{-1/2}$ and $K' = K$. We then get $$a(q) = -\frac{\log 2}{8} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{24}$$ and $$a(q^{2}) = -\frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{12}$$ and $$a(q^{4}) = \frac{3\log 2}{8} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{6}$$ so we get $$72a(q) - 96a(q^{2}) + 24a(q^{4}) = -3\pi + 8\pi - 4\pi = \pi$$ Notice that the other terms cancel very nicely.

Note: The above derivation requires some knowledge of elliptic integrals and their connection with the theta functions. The technique of going from $q$ to $q^{2}$ is the famous Landen Transformation. Material related to these nice theories is presented in my blog and the interested reader may search elliptic integral, theta functions and Landen in the archives page.

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