complex-analysissequences-and-series
Consider the sum
$$\frac{3 \sqrt{3}}{2\pi}\sum_{z\in \Lambda} \frac{1}{1-\left(\dfrac{z}{\sqrt{3}}-1\right)^3} \overset{?}=1$$
with $\Lambda=3\mathbb Ze^{\pi i/6}+3\mathbb Ze^{-\pi i/6}$, then it is numerically not too difficult to see that this sum is equal to $1.$ I am however looking for an analytic argument of this nice fact. I assume it must rely on some subtle symmetries?
Please let me know if you have any questions.
Let $\alpha = a$, $$c_n = \frac{\Gamma\left(\frac{n+\alpha+1}{2}\right)}{\Gamma\left(\frac{n+\alpha+2}{2}\right)}$$ using the fact that $\Gamma(x+1) = x\Gamma(x)$, then $$c_{n+2} = \frac{n+\alpha+1}{n+\alpha+2}c_n$$
Let
$$f(x) = \sum_{n=0}^\infty (-1)^n c_n x^n$$
then,
$$f'(x) = \sum_{n=0}^\infty n(-1)^n c_n x^{n-1}$$
Start by proving that: $$\left(1-x^2\right) xf'(x) + \left(\alpha(1 - x^2) - x^2\right)f(x) = \beta - \gamma x$$ with \begin{align} \beta &= \alpha c_0\\ \gamma &= (\alpha + 1)c_1. \end{align} Indeed, \begin{align} \left(1-x^2\right) xf'(x) + \left(\alpha(1 - x^2) - x^2\right)f(x) &= \sum_{n=0}^{\infty} n(-1)^nc_nx^n - \sum_{n=0}^{\infty} n(-1)^nc_nx^{n+2} + \sum_{n=0}^{\infty} \alpha (-1)^n c_nx^n - \sum_{n=0}^{\infty} (\alpha+1) (-1)^n c_nx^{n+2}\\ &= \alpha c_0 -(\alpha+1)c_1 x + \sum_{n=0}^{\infty} (-1)^n \underbrace{\left((n+\alpha +2)c_{n+2} - (n+\alpha+1)c_n\right)}_{=0} x^{n+2}\\ &= \beta - \gamma x \end{align}
Let $g(x) = f(x)x^\alpha \sqrt{1 - x^2}$, then \begin{align} g'(x) &= f'(x) x^{\alpha}\sqrt{1-x^2} + \alpha f(x) x^{\alpha - 1}\sqrt{1-x^2} - f(x)x^{\alpha}\frac{x}{\sqrt{1-x^2}}\\ &= \frac{x^{\alpha - 1}}{\sqrt{1 - x^2}}\left(\left(1-x^2\right) xf'(x) + \left(\alpha(1 - x^2) - x^2\right)f(x)\right)\\ &=\frac{x^{\alpha - 1}}{\sqrt{1 - x^2}}\left(\beta - \gamma x\right) \end{align}
L'Hôpital's rule: since $g(1)=0$, \begin{align} f(1) &=\lim_{x\to 1} \frac{g(x)}{x^\alpha \sqrt{1-x^2}}\\ &=\lim_{x\to 1} \frac{g'(x)}{\displaystyle\alpha x^{\alpha-1}\sqrt{1-x^2} - x^{\alpha} \frac{x}{\sqrt{1-x^2}}}\\ &= \lim_{x\to 1} \frac{\displaystyle\left(\beta -\gamma x\right) \frac{x^{\alpha -1}}{\sqrt{1-x^2}}}{\displaystyle\alpha x^{\alpha-1}\sqrt{1-x^2} - x^{\alpha} \frac{x}{\sqrt{1-x^2}}}\\ &= \lim_{x\to 1} \frac{\displaystyle\left(\beta - \gamma x\right)}{\displaystyle\alpha \left(1-x^2\right) - x^2}\\ &= \gamma - \beta. \end{align}
To finish the proof you need to simplify $\gamma - \beta$ which is not difficult.
Let us use the more standard notation with $k\in(0,1)$ being elliptic modulus and $k'=\sqrt{1-k^2}$ being the complementary modulus. The elliptic integrals are defined by $$K(k) =\int_0^{\pi/2}\frac {dx} {\sqrt{1-k^2\sin^2x}},E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2x}\,dx\tag{1}$$ The integrals $K(k), K(k'), E(k), E(k') $ are often denoted by $K, K', E, E'$ when the values $k, k'$ are available from context.
Let us now bring other players into this game and we write $q=\exp(-\pi K'/K) $ and call it the nome corresponding to modulus $k$. Ramanujan now makes use of the Dedekind eta function $$\eta(q) =q^{1/24}\prod_{n\geq 1}(1-q^n)\tag{2}$$ and its logarithmic derivative $$P(q) = 24q\frac{d}{dq}\log\eta(q)=1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n}\tag{3}$$ We have the identity $$\eta(q^2)=2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{4}$$ which will be used in combination with the hypergeometric identity $$ \left(\frac{2K}{\pi}\right)^{2} = (1 - 2k^{2})^{-1}\,_{3}F_{2}\left(\frac{1}{4},\frac{3}{4},\frac{1}{2}; 1, 1; -\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2}\right)\tag{5}$$ where $G=G(k) =(2kk') ^{-1/12}$ is one of Ramanujan's class invariant. Using these identities we get $$\eta^4(q^2)=\frac{2^{-4/3}(kk')^{2/3}}{1-2k^2} \,_{3}F_{2}\left(\frac{1}{4},\frac{3}{4},\frac{1}{2}; 1, 1; -\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2}\right)\tag{6}$$ Next formidable step is to logarithmically differentiate the above identity with respect to $k$ to get a formula for $P(q^2)$. The calculations can be verified with hand and some patience and we write a few steps here. First we have $$P(q^2)=3q\frac{d}{dq}\log\eta^4(q^2)=\frac{3kk'^2}{2}\left(\frac{2K}{\pi}\right)^2\frac{d}{dk}\log\eta^4(q^2)$$ which further leads us to $$P(q^2)=\frac{3kk'^2}{2}\left(\frac{2K}{\pi}\right)^2\frac{d}{dk}\left(\log (2^{-4/3}(kk')^{2/3})+\log\left(\frac {2K}{\pi}\right) ^2\right) $$ Then we get $$P(q^2)=\left(\frac{2K}{\pi}\right) ^2(1-2k^2)+\frac{3kk'^2}{2}\frac{d}{dk}\left(\frac{2K}{\pi}\right)^2$$ Using $(5)$ we now get $$P(q^2)=\left(\frac{2K}{\pi}\right) ^2\cdot\frac{1+2k^2k'^2}{1-2k^2}+\frac{3kk'^2}{2(1-2k^2)}\frac{d}{dk}\,_{3}F_{2}\left(\frac{1}{4},\frac{3}{4},\frac{1}{2}; 1, 1; -\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2}\right)$$ Switching to sigma notation we get $$P(q^2)=\frac{1+2k^2k'^2}{(1-2k^2)^2}\sum_{n\geq 0}(-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2n}+\frac{3kk'^2}{2(1-2k^2)}\cdot\frac{f'(k)}{f(k)}\sum_{n\geq 0}(-1)^n\cdot\frac{n(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2n}$$ where $$f(k)= \left(\frac{G^{12} - G^{-12}}{2}\right)^{-2}$$ We have $$\frac{f'(k)} {f(k)} =(\log f(k)) '=\frac{2(1-2k^2)}{kk'^2}\cdot\frac{G^{12}+G^{-12}}{G^{12}-G^{-12}}$$ Thus we finally obtain $$P(q^2)=\frac{1+2k^2k'^2}{(1-2k^2)^2}\sum_{n\geq 0}(-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2n}+\frac{3(G^{12}+G^{-12})}{G^{12}-G^{-12}}\sum_{n\geq 0}(-1)^n\cdot\frac{n(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2n}\tag{7} $$ Next we put $q=e^{-5\pi}$ in above relation and corresponding values $$2k^2=1-\sqrt{1-\phi^{-24}},2k'^2=1+\sqrt{1-\phi^{-24}}, G=(4k^2k'^2)^{-1/24}=\phi$$ (where $\phi=(1+\sqrt{5})/2$ is golden ratio) and observe that $$G^{12}-G^{-12}=F_{12}\phi+F_{11}-(-F_{12}\phi+F_{13})=2F_{12}\phi-F_{12}=F_{12}\sqrt{5}=144\sqrt{5}$$ (where $F_n$ is the Fibonacci sequence) and $$G^{12}+G^{-12}=F_{11}+F_{13}=322$$ and $$\frac{1+2k^2k'^2}{(1-2k^2)^2}=\frac{2F_{24}\phi+2F_{23}+1}{2F_{24}\phi+2F_{23}-2}=\frac{30912\phi+19105}{30912\phi+19104}$$ The above ratio can be simplified (using technique in this answer) to $$\frac{120+161\sqrt{5}}{480}$$ Thus the equation $(7)$ can be rewritten as $$P(q^2)=\frac{120+161\sqrt{5}}{480}\sum_{n\geq 0}(-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\frac{1}{25920^n}+\frac{644\sqrt{5}}{480}\sum_{n\geq 0}(-1)^n\cdot\frac{n(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\frac{1}{25920^n}\tag{8}$$ From equation $(15)$ of this answer dealing with modular equations and related identities we have $$P(q^{2})=\frac{3}{5\pi}+\left(\frac{2K}{\pi}\right) ^26(5-3\phi)\sqrt{3\phi-4}$$ or $$ P(q^{2})=\frac{3}{5\pi}+\frac{6(5-3\phi)\sqrt{3\phi-4}} {\sqrt{1-\phi^{-24}}} \sum_{n\geq 0} (-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\frac{1}{25920^n}$$ and this simplifies to $$P(q^{2})=\frac{3}{5\pi}+\frac{120+120\sqrt{5}} {480} \sum_{n\geq 0} (-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\frac{1}{25920^n}\tag{9}$$ Comparing equations $(8),(9)$ we get $$\frac{288\sqrt{5}}{5\pi}=\sum_{n\geq 0}(-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\frac{41+644n}{25920^n} $$
Best Answer
Let $\Lambda' = (e^{i\pi/6}\mathbb{Z} \oplus e^{-i\pi/6}\mathbb{Z}) - \frac{1}{\sqrt{3}}$. Then by substituting $z = 3(\omega + \frac{1}{\sqrt{3}})$,
$$ S := \sum_{z \in \Lambda} \frac{1}{1-((z/\sqrt{3})-1)^3} = \sum_{\omega \in \Lambda'} \frac{1}{1-(\sqrt{3} \omega)^3}. $$
Our goal is to show that
Now by noting that $\Lambda'$ can be decomposed into concentric "discrete regular triangles" centered at the origin, let $\Lambda'_N$ be the union of the first $N$ smallest triangles. For example, $\Lambda'_5$ is
In particular, each $\Lambda'_N$ is symmetric about a $\frac{2\pi}{3}$-rotation about the origin. Then by using the partial fraction decomposition
$$ \frac{1}{1 - w^3} = \frac{1}{3} \sum_{\xi \, : \, \xi^3 = 1} \frac{1}{1 - \xi w}, $$
we find that
\begin{align*} S = \lim_{N\to\infty} \sum_{\omega \in \Lambda'_N} \frac{1}{1-(\sqrt{3} \omega)^3} = \lim_{N\to\infty} \sum_{\omega \in \Lambda'_N} \frac{1}{1-\sqrt{3}\omega} = \frac{1}{\sqrt{3}} \lim_{N\to\infty} \sum_{\tau \in \Lambda''_N} \frac{1}{\tau}, \end{align*}
where $\tau = \frac{1}{\sqrt{3}} - \omega$ and $\Lambda''_N = \frac{1}{\sqrt{3}} - \Lambda'_N$. For example, $\Lambda''_5$ is:
Now by regrouping the dots in $\Lambda''_N$ into concentric regular triangles centered at the origin, we are left with $N-1$ triangles plus two extra "discrete lines", which we denote by $\gamma_N$. For example, the next picture demonstrates the decomposition of $\Lambda''_5$ into the triangles and $\gamma_5$:
Since the sum along each concentric triangle vanishes by symmetry, we are left with
\begin{align*} S = \frac{1}{\sqrt{3}} \lim_{N\to\infty} \sum_{\tau \in \gamma_N} \frac{1}{\tau} = \frac{1}{\sqrt{3}} \lim_{N\to\infty} \sum_{\tau \in \gamma_N} \frac{1}{(\tau/N)} \cdot \frac{1}{N} \end{align*}
Now the last one can be recognized as a Riemann sum for a contour integral. Indeed, if we set
$$ z_1 = -\frac{\sqrt{3}}{2} - \frac{3i}{2}, \qquad z_2 = \sqrt{3}, \qquad z_3 = -\frac{\sqrt{3}}{2} + \frac{3i}{2} $$
so that the polygonal line $\overline{z_1 z_2} \cup \overline{z_2 z_3}$ is the "limit of the rescaled discrete line $N^{-1}\gamma_N$" as $N\to\infty$, then
\begin{align*} S &= \frac{1}{\sqrt{3}} \int_{\overline{z_1 z_2} \cup \overline{z_1 z_2}} \frac{|\mathrm{d}z|}{z} \\ &= \frac{1}{\sqrt{3}} \biggl( \frac{1}{e^{i\pi/6}} \int_{[z_1, z_2]} \frac{\mathrm{d}z}{z} + \frac{1}{e^{i5\pi/6}} \int_{[z_2, z_3]} \frac{\mathrm{d}z}{z} \biggr) \\ &= \frac{1}{\sqrt{3}} \biggl( \frac{1}{e^{i\pi/6}} \cdot \frac{2\pi i}{3} + \frac{1}{e^{i5\pi/6}} \cdot \frac{2\pi i}{3} \biggr) \\ &= \frac{2\pi}{3\sqrt{3}}. \end{align*}
Therefore the desired conclusion follows.