\documentclass[11pt]{amsart} \usepackage[T1]{fontenc} \usepackage{lmodern,microtype,amsmath,amssymb,amsthm,mathtools,booktabs} \usepackage[margin=1in]{geometry} \usepackage[hidelinks]{hyperref} \usepackage{xurl} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{remark}\newtheorem{remark}[theorem]{Remark} \newcommand{\tht}{\vartheta} \newcommand{\F}{\mathcal F} \newcommand{\ii}{\mathrm i} \title{Three cubic trinomial series for $1/\pi$} \author{Khai-Hoan Nguyen-Dang} \subjclass[2020]{Primary 33C20; Secondary 11F03, 33C45.} \keywords{Ramanujan-type series, generalized central trinomial coefficients, Legendre polynomials, elliptic integrals, modular equations} \date{29 October 2026} \begin{document} \begin{abstract} We prove the three identities in Sun's Conjecture VI, involving cubes of generalized central trinomial coefficients. A common double integral represents their generating functions. On two rational curves, this period reduces to quadratic expressions in Gauss hypergeometric functions associated with elliptic integrals. A first-order identity recovers the derivative transverse to each curve and supplies the linear weights in the three series. The constants are determined by an endpoint calculation and Bauer's identity. Classical modular equations of degrees three, five, and seven then complete the evaluations. We give explicit rational differential certificates for the period reductions and the transverse derivative identities. \end{abstract} \maketitle \section{Introduction} We prove the three cubic trinomial identities for $1/\pi$ in Sun's Conjecture VI \cite[Section~5]{SunChapter}. The main step is a pair of one-parameter transformations for a common two-variable period, together with identities for its transverse derivatives. These formulas connect the three series to classical elliptic periods and modular equations. For $b,c\in\mathbb C$ and an integer $n\geq0$, define \begin{equation}\label{def:T} T_n(b,c)=[X^n](X^2+bX+c)^n =\sum_{j=0}^{\lfloor n/2\rfloor} \binom n{2j}\binom{2j}j b^{n-2j}c^j. \end{equation} Sun recorded the discovery of the three formulas on 1 and 2 October 2011 \cite[Section~4]{SunList} and published them together as Conjecture VI in \cite{SunChapter}. He still described all three type-VI series as open in 2020 \cite[p.~1275]{Sun2020}. Our main result proves the complete conjecture. \begin{theorem}\label{thm:main} The following absolutely convergent identities hold: \begin{align} \sum_{n=0}^\infty (66n+17)\frac{T_n(10,121)^3}{55296^n} &=\frac{540\sqrt2}{11\pi},\label{eq:VI1}\\ \sum_{n=0}^\infty (126n+31) \frac{T_n(22,441)^3}{(-512000)^n} &=\frac{880\sqrt5}{21\pi},\label{eq:VI2}\\ \sum_{n=0}^\infty (3990n+1147) \frac{T_n(62,9025)^3}{(-23887872)^n} &=\frac{432(195\sqrt{14}+94\sqrt2)}{95\pi}.\label{eq:VI3} \end{align} \end{theorem} Ramanujan's 1914 paper \cite{Ramanujan} placed series for $1/\pi$ in the setting of elliptic integrals and modular equations. The subsequent theory includes transformations between different hypergeometric descriptions of elliptic periods; the work of Berndt, Bhargava, and Garvan \cite{BBG} develops the alternative bases used below. Sun's formulas bring generalized central trinomial coefficients into this setting. Their relation to Legendre polynomials is \begin{equation}\label{eq:LegendreT} T_n(b,c)=(b^2-4c)^{n/2} P_n\left(\frac b{\sqrt{b^2-4c}}\right). \end{equation} Here $P_n$ is the Legendre polynomial normalized by $P_n(1)=1$. For $b^2\ne4c$, the expression is independent of the square root; after expansion it is a polynomial, which also gives its value when $b^2=4c$. Thus the problem belongs to the study of generating functions of powers of Legendre polynomials. Zudilin \cite{ZudilinSquares} obtained a transformation for a binomially weighted generating function of Legendre squares and used it to prove a family of Sun's series. Van Hoeij, van Straten, and Zudilin \cite[Section~6]{Saga} later gave a Hadamard-product representation for the generating function of Legendre cubes and studied its fourth-order differential equation. That work provides a general framework for the cubic problem. The reductions here concern two particular rational curves: on each, the relevant period satisfies a third-order equation that is a symmetric square of a second-order equation. For the second curve, the resulting equation is related to the hypergeometric transformation for Domb numbers in Rogers \cite[Theorem~3.1, equation~(3.4)]{Rogers}. There is a further derivative to determine. If $F(w)=\sum a_nw^n$, then a weight $An+B$ corresponds to applying $Aw\,d/dw+B$ to $F$. In our two-variable generating function, this derivative holds the trinomial parameter fixed, whereas restricting to a curve gives a tangent derivative. Differentiation and Legendre's relation have already been used systematically to obtain series for $1/\pi$ from products of elliptic periods; see Wan \cite[Section~2]{Wan}. Here we recover the missing direction through a first-order identity. Its constant is fixed analytically, by an endpoint estimate on the first curve and Bauer's identity on the second. The resulting formulas apply along the entire curves; the three series follow by evaluation at the required algebraic points. The proof has three parts. Section~\ref{sec:integral} gives the double-integral representation. Sections~\ref{sec:elliptic}--\ref{sec:first-evaluations} establish the first period transformation, its transverse derivative, and the first and third series. Section~\ref{sec:second} proves the second series using a signature-three transformation and an explicit continuation path. Appendices~\ref{app:certificate} and~\ref{app:branches} contain the rational differential certificates and the polynomial sign checks for that path. All square roots and continuation branches used in the proof are specified. \section{An integral and a rational curve}\label{sec:integral} Use the normalization \[ G(h,w)=\sum_{n=0}^\infty T_n(h,1/4)^3w^n, \qquad \theta_w=w\frac{\partial}{\partial w}. \] Since \[ T_n(h,1/4)=\frac1\pi\int_0^\pi(h+\cos v)^n\,dv, \] uniform convergence of the geometric series gives \begin{equation}\label{eq:triple} G(h,w)=\frac1{\pi^3}\int_{[0,\pi]^3} \frac{dv_1\,dv_2\,dv_3} {1-w\prod_{j=1}^3(h+\cos v_j)} \end{equation} whenever $|w|(|h|+1)^3<1$. The integral also defines analytic continuation on real parameter intervals where its denominator remains positive. Define \begin{equation}\label{eq:Psi} \Psi(s,t)=\frac1{\pi^2}\int_0^1\int_0^1 \frac{du\,dv} {\sqrt{u(1-u)v(1-v)(1-suv)(1-t(1-u)(1-v))}}. \end{equation} For real $s,t<1$, all square roots in this formula are positive. Expanding the last two factors and evaluating beta integrals proves \begin{equation}\label{eq:PsiSeries} \Psi(s,t)=\sum_{m,n\geq0} \frac{(1/2)_m^3(1/2)_n^3}{m!n!(m+n)!^2}s^m t^n \qquad(|s|,|t|<1). \end{equation} Here $(a)_n=a(a+1)\cdots(a+n-1)$ and $(a)_0=1$. The substitution $(u,v)\mapsto(1-u,1-v)$ gives $\Psi(s,t)=\Psi(t,s)$. \begin{lemma}\label{lem:mobius} Let \[ h=-\frac{a+b}{1+ab},\qquad w=\frac{a}{(h^2-1)(1+ah)}, \] and put \[ s=\frac{4a}{(1-b)(1+a)},\qquad t=-\frac{4a}{(1+b)(1-a)}. \] Suppose $a,b\in(-1,1)$, the displayed expressions are defined, and $s,t<1$. Then, with $G$ represented by \eqref{eq:triple}, \begin{equation}\label{eq:mobius} G(h,w)=\frac{\sqrt{1-a^2}}{1+ab}\Psi(s,t). \end{equation} In particular this identity holds on the curve \begin{align} b&=-\frac{5a+1}{a+1},& h&=\frac{a^2-4a-1}{5a^2-1},\notag\\ w&=-\frac{(5a^2-1)^3} {8(a-1)^2(a+1)^3(3a+1)},& -\frac13&|B|$. Thus the integral in $Z$ equals $(A^2-B^2)^{-1/2}$. Moreover $d\nu(X)$ and $d\nu(Y)$ become the beta measures $du/(\pi\sqrt{u(1-u)})$ and $dv/(\pi\sqrt{v(1-v)})$. The constant extracted from $\sqrt{A^2-B^2}$ is $\sqrt{(1-a^2)(1-b^2)}$, which proves precisely the prefactor in \eqref{eq:mobius}. This argument also proves positivity of the original denominator, since all the linear factors in the substitution are positive. For \eqref{eq:curve}, substitution gives $s=2a/(3a+1)<0$ and $t=(a+1)/(1-a)\in(1/2,1)$. Also $|a|,|b|<1$ there. Hence the same direct integral argument proves \eqref{eq:gPsi} on the entire stated interval. \end{proof} Set \begin{equation}\label{eq:u} U=\frac{1-a}{1+3a}>1. \end{equation} The curve in the $(s,t)$-plane is $s+t-2st=1$, parametrized by $s=(1-U)/2$ and $t=(1+U^{-1})/2$. \section{The elliptic transformation}\label{sec:elliptic} Write \[ f(x)={}_2F_1(1/2,1/2;1;x),\qquad H(x)={}_3F_2(1/2,1/2,1/2;1,1;x). \] Thus $f(x)=2K(x)/\pi$, where $K(x)=\int_0^{\pi/2}(1-x\sin^2 v)^{-1/2}\,dv$ uses the parameter convention. Our hypergeometric normalization is \[ {}_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;x) =\sum_{n\geq0}\frac{(a_1)_n\cdots(a_p)_n}{(b_1)_n\cdots(b_q)_n} \frac{x^n}{n!}, \] initially in its disk of convergence, and continued analytically where stated. We use Clausen's identity, Pfaff's transformation, and Legendre's relation in the forms recorded in \cite[16.12.2, 15.8.1, 19.7.1]{DLMF}; the quadratic forms used below also follow by substitution in the Gauss equation. \begin{theorem}\label{thm:period} For $U>1$, put \[ m=\frac{1-\sqrt{1-U^{-2}}}{2}. \] Then \begin{equation}\label{eq:period} \Psi\left(\frac{1-U}{2},\frac{1+U^{-1}}2\right) =\frac{3f(m)^2+f(1-m)^2}{4\sqrt U}. \end{equation} \end{theorem} The coefficient $3$ in this formula is fixed by the local behavior at $U=1$; it cannot be inferred from the differential equation alone. \begin{lemma}\label{lem:ODE} The function on the left of \eqref{eq:period} is annihilated by \begin{equation}\label{eq:ODE} (2\tht-1)^3-U^2(2\tht+1)^3, \qquad \tht=U\frac{d}{dU}. \end{equation} \end{lemma} \begin{proof} The equation follows by exact elimination from the Legendre recurrence. The derivation and explicit certificate are given in Appendix~\ref{app:certificate}. All identities used in that elimination are over a rational-function field. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:period}] Let $f_0=f(1/2)$ and $f_1=f'(1/2)$. The Gauss equation and Legendre's relation give \begin{equation}\label{eq:midpoint} f''(1/2)=f_0,\qquad f_0f_1=2/\pi. \end{equation} Put $J(t)={}_3F_2(1/2,1/2,1/2;2,2;t)$. Coefficient comparison in the defining series gives the contiguous identity \begin{equation}\label{eq:contiguous} J(t)=8H(t)+(32-8/t)\theta_tH(t) +48(1-1/t)\theta_t^2H(t). \end{equation} The apparent singularity at $t=0$ is removable. The quadratic Clausen identity is \begin{equation}\label{eq:Clausen} H(t)=f\left(\frac{1-\sqrt{1-t}}2\right)^2. \end{equation} It follows from \eqref{eq:midpoint} that, with $r=\sqrt{1-t}$, \[ H(t)=f_0^2-\frac2\pi r+\frac{f_1^2+f_0^2}{4}r^2+O(r^3). \] Substitution in \eqref{eq:contiguous} cancels the terms in $r^{-1}$ and gives \begin{equation}\label{eq:Jone} J(1)=2f_0^2-6f_1^2. \end{equation} From \eqref{eq:PsiSeries}, uniformly for $0\leq t\leq1$ and small real $s$, \[ \Psi(s,t)=H(t)+\frac{s}{8}J(t)+O(s^2). \] For the uniform bound, the coefficient in \eqref{eq:PsiSeries} is at most the product of the corresponding coefficients of $H(s)H(t)$, and $H(1)$ is finite. Moreover $J'(1)$ is finite by absolute convergence of its differentiated series, so $J(t)=J(1)+O(1-t)$. Consequently, as $\delta\downarrow0$, \begin{equation}\label{eq:initial} \Psi\left(-\frac\delta2,\frac{1+(1+\delta)^{-1}}2\right) =f_0^2-\frac{\sqrt2}{\pi}\sqrt\delta +\frac{f_1^2}{2}\delta+O(\delta^{3/2}). \end{equation} To identify a local basis, put \[ u(U)=U^{-1/4}f(m),\qquad v(U)=U^{-1/4}f(1-m). \] The Gauss equation, followed by the change of variable $4m(1-m)=U^{-2}$, shows that both functions satisfy \[ Y''+\frac{3U^2-1}{2U(U^2-1)}Y' +\frac{1}{16U^2}Y=0. \] Its symmetric square is exactly~\eqref{eq:ODE}, after division by its leading coefficient. Hence $u^2,uv,v^2$ solve that equation. As $\delta=U-1\downarrow0$, the three combinations \[ uv,\qquad v^2-u^2,\qquad (v-u)^2 \] have leading terms \[ f_0^2,\qquad 2\sqrt2\,f_0f_1\sqrt\delta,\qquad 2f_1^2\delta, \] respectively, all with nonzero coefficients. They therefore form a local basis. The right side of~\eqref{eq:period} is $(3u^2+v^2)/4$ and has the three coefficients in~\eqref{eq:initial}. Comparison in this basis proves the identity near $U=1$. Since the leading coefficient of~\eqref{eq:ODE} does not vanish on $10$. \begin{proposition}\label{thm:normal} With $g,h,w$ as in \eqref{eq:curve}--\eqref{eq:gPsi}, \begin{equation}\label{eq:normal} (\theta_wG)(h(a),w(a))+A_0(a)g(a)+A_1(a)g'(a) =\frac{3\sqrt2}{8\pi}R(a). \end{equation} \end{proposition} \begin{proof} Set \[ N(a)=(\theta_wG)(h(a),w(a)),\qquad E=N+A_0g+A_1g'. \] The normal row identity in Appendix~\ref{app:certificate} gives $E'=\rho E$, where direct differentiation shows that $\rho=R'/R$. Hence $E/R$ is constant. An endpoint calculation determines the constant. As $a\to0^-$, \[ h=1+4a+O(a^2),\quad w=\tfrac18(1-4a+O(a^2)), \quad q:=w(h+1)^3=1+2a+O(a^2). \] The elementary endpoint estimate in the single cosine integral is \[ T_n(h,1/4)=(h+1)^n\sqrt{\frac{h+1}{2\pi n}} (1+O(n^{-1})), \] uniformly for $h$ in a sufficiently small neighborhood of $1$. The analogous differentiated estimate follows by inserting $n(h+\cos v)^{n-1}$ in the same integral. Expanding $\cos v=1-v^2/2+O(v^4)$ near zero and bounding the remaining interval geometrically proves both estimates. Cubing and summing gives \[ N\sim\frac1{\pi\sqrt{1-q}},\qquad hG_h\sim\frac{3h}{h+1}N,\qquad g=O(1). \] Here the error in the differentiated sum is bounded by a convergent multiple of $\sum n^{-3/2}$, and $\sum_{n\geq1}q^n n^{-1/2}\sim\sqrt\pi/\sqrt{1-q}$ follows by a rescaled Riemann sum. Since $h'/h\to4$ and $w'/w\to-4$, we have $g'\sim2N$. Finally, \[ A_0\to\frac38,\quad A_1\to-\frac18, \quad R\sim(-a)^{-1/2}, \] so $E/R\to3\sqrt2/(8\pi)$. This proves \eqref{eq:normal} on a neighborhood of zero and hence throughout the interval by its first-order equation and analytic continuation of the positive integral. \end{proof} \section{Classical modular evaluations}\label{sec:CM} For $0<\lambda<1$, put \[ F_\lambda(u)={}_2F_1(\lambda,1-\lambda;1;u),\qquad H_\lambda(x)={}_3F_2(1/2,\lambda,1-\lambda;1,1;x),\qquad \sigma_\lambda=\frac{\sin(\pi\lambda)}{\pi}. \] The quadratic transformation and Clausen's identity give \begin{equation}\label{eq:general-quadratic-Clausen} H_\lambda(4u(1-u))=F_\lambda(u)^2\qquad(01$, and let $m\in(0,1/2)$ be the unique point such that \[ \mathcal R_\lambda(m)=\sqrt N, \qquad \mathcal R_\lambda(u)=\frac{F_\lambda(1-u)}{F_\lambda(u)}. \] Write $p=m(1-m)$ and $d=1-2m$. Let $\beta(\alpha)$ be the local branch through $(m,1-m)$ determined by $\mathcal R_\lambda(\beta)=\mathcal R_\lambda(\alpha)/N$. Then $\beta'(m)=1$ and \begin{equation}\label{eq:general-modular-derivative} \left[\theta_x H_\lambda(x)\right]_{x=4p} +\left(1+\frac{p\beta''(m)}{2d}\right)H_\lambda(4p) =\frac{\sigma_\lambda}{d\sqrt N}, \qquad \theta_x=x\frac{d}{dx}. \end{equation} \end{lemma} \begin{proof} Both $F_\lambda(u)$ and $F_\lambda(1-u)$ solve the Gauss equation \[ u(1-u)Y''+(1-2u)Y'-\lambda(1-\lambda)Y=0. \] Abel's identity makes their Wronskian a constant multiple of $1/(u(1-u))$. Euler's integral, and the same integral differentiated in $u$, give \[ F_\lambda(1-u)=-\sigma_\lambda\log u+O(1),\qquad \frac{d}{du}F_\lambda(1-u)\sim-\frac{\sigma_\lambda}{u} \quad(u\downarrow0). \] Since $F_\lambda(u)=1+O(u)$, the Wronskian constant is $-\sigma_\lambda$. This proves~\eqref{eq:general-wronskian}; its midpoint specialization gives~\eqref{eq:general-midpoint}. It also gives \begin{equation}\label{eq:general-ratio-derivative} \mathcal R_\lambda'(u) =-\frac{\sigma_\lambda}{u(1-u)F_\lambda(u)^2}<0. \end{equation} Together with $\mathcal R_\lambda(1/2)=1$ and $\mathcal R_\lambda(u)\to\infty$ as $u\downarrow0$, this proves existence and uniqueness of $m$. At the complementary pair, $F_\lambda(1-m)=\sqrt N F_\lambda(m)$, so~\eqref{eq:general-ratio-derivative} gives $\beta'(m)=1$. Set \[ A=\frac{F_\lambda'(m)}{F_\lambda(m)},\qquad B=\frac{F_\lambda'(1-m)}{F_\lambda(1-m)}. \] Equation~\eqref{eq:general-wronskian} and logarithmic differentiation of the defining equation for $\beta$ give, respectively, \[ A+B=\frac{\sigma_\lambda}{p\sqrt N F_\lambda(m)^2},\qquad \beta''(m)=-\frac{2d}{p}+2(B-A). \] Finally, differentiation of~\eqref{eq:general-quadratic-Clausen} gives \[ \left[\theta_xH_\lambda(x)\right]_{x=4p} =\frac{2p}{d}A F_\lambda(m)^2. \] Eliminating $A,B$ proves~\eqref{eq:general-modular-derivative}. \end{proof} For $N=3$ or $7$, put $\gamma=1/(N+1)$. The two modular equations needed are \begin{equation}\label{eq:modular} (\alpha\beta)^\gamma+((1-\alpha)(1-\beta))^\gamma=1. \end{equation} Here $\alpha,\beta\in(0,1)$ satisfy $\mathcal R_{1/2}(\beta)=N\mathcal R_{1/2}(\alpha)$, and all roots are positive. For $N=3$ this is \cite[(1.2), p.~31]{ChanLiaw}; for $N=7$ it is \cite[Example 2, p.~39]{ChanLiaw}. Their convention for degree is stated in their equation (1.1). \begin{lemma}\label{lem:CM} At \[ m_3=\frac{2-\sqrt3}{4},\qquad m_7=\frac{8-3\sqrt7}{16}, \] one has $f(1-m_N)=\sqrt N f(m_N)$. Moreover, \begin{equation}\label{eq:Ramanujan} (6\theta+1)H(1/4)=\frac4\pi, \qquad (42\theta+5)H(1/64)=\frac{16}\pi, \qquad\theta=x\frac{d}{dx}. \end{equation} Operators in this notation are applied before evaluation. \end{lemma} \begin{proof} Use Lemma~\ref{lem:modular-differentiation} with $\lambda=1/2$. The unique $m\in(0,1/2)$ with $\mathcal R_{1/2}(m)=\sqrt N$ forms a complementary degree-$N$ pair $(1-m,m)$. Equation~\eqref{eq:modular} gives $2\{m(1-m)\}^{\gamma}=1$, which yields the stated values $m_N$. For the inverse local branch through $(m,1-m)$, put $p=m(1-m)$ and $d=1-2m$. Twice differentiating the symmetric modular equation gives \[ \beta'=1,\qquad \beta''=\frac{2(\gamma-1)+4p}{dp}. \] Equation~\eqref{eq:general-modular-derivative} consequently becomes \begin{equation}\label{eq:CMuniform} \theta H(4p)+\frac{\gamma-2p}{1-4p}H(4p) =\frac1{\pi d\sqrt N}. \end{equation} At $N=3$, $p=1/16$ and $d=\sqrt3/2$; at $N=7$, $p=1/256$ and $d=3\sqrt7/8$. These substitutions give~\eqref{eq:Ramanujan}. \end{proof} \section{Evaluation of the first and third series}\label{sec:first-evaluations} \begin{proof}[Proof of \eqref{eq:VI1} and \eqref{eq:VI3}] On the curve put \[ W(a)=\frac{P(a)}{4\sqrt{U(a)}},\qquad m(a)=\frac{1-\sqrt{1-U(a)^{-2}}}{2}. \] Theorem~\ref{thm:period} says $g=W(3f(m)^2+f(1-m)^2)$. At the two points needed here, direct substitution gives the following exact data: \[ \begin{array}{c|cc} &a=-1/7&a=-7/25\\\hline U&2&8\\ P&6\sqrt3/11&108/95\\ A_0&65/242&2509/10830\\ A_1&-8/539&1504/59375\\ R&240/121&7488\sqrt7/12635 \end{array} \] Let $V=f(m)^2$ at the relevant point. Differentiation, using Lemma~\ref{lem:CM} and the complementary derivative identities in the proof of Lemma~\ref{lem:modular-differentiation}, yields \begin{align*} g(-1/7)&=\frac{9\sqrt6}{22}V,& g'(-1/7)&=\frac{147\sqrt6}{484}V,\\ g(-7/25)&=\frac{27\sqrt2}{38}V,& g'(-7/25)&=\frac{1125\sqrt2}{161728} \left(225V-\frac{608}{\pi}\right). \end{align*} The derivatives are taken in the period identity before specializing the parameter. For any constants $A,B$, Proposition~\ref{thm:normal} gives \begin{equation}\label{eq:weighted} (A\theta_w+B)G =\frac{3A\sqrt2}{8\pi}R+(B-AA_0)g-AA_1g'. \end{equation} At $a=-1/7$, insert $(A,B)=(66,17)$. The coefficient of $V$ is zero, and the right side is $540\sqrt2/(11\pi)$. At $a=-7/25$, insert $(A,B)=(3990,1147)$. Again the coefficient of $V$ is zero, and the result is \[ \frac{40608\sqrt2}{95\pi}+\frac{16848\sqrt{14}}{19\pi} =\frac{432(94\sqrt2+195\sqrt{14})}{95\pi}. \] To identify the original series, use the scaling identity $T_n(b,c)=(2\sqrt c)^nT_n(b/(2\sqrt c),1/4)$. At the first point, $h=5/11$ and $w=1331/6912=22^3/55296$. At the second point, $h=-31/95$ and $w=857375/2985984$; the parity identity $T_n(-h,1/4)=(-1)^nT_n(h,1/4)$ supplies the negative denominator in \eqref{eq:VI3}. Finally, $|T_n(b,c)|\leq(|b|+2\sqrt c)^n$ proves absolute convergence: the respective geometric bounds are $16/27$ and $343/512$. It also justifies termwise application of the linear weight. \end{proof} \section{The second identity}\label{sec:second} The second target lies on $s+t=-1$, rather than on the curve of Theorem~\ref{thm:period}. We determine its period and its transverse derivative separately. Write \[ f_3(u)={}_2F_1(1/3,2/3;1;u),\qquad H_3(v)={}_3F_2(1/2,1/3,2/3;1,1;v). \] Thus $f_3=F_{1/3}$ and $H_3=H_{1/3}$ in the notation of Section~\ref{sec:CM}. The branch of each hypergeometric function is the one continued from its power series at zero to the plane cut along $[1,\infty)$. \begin{samepage} \begin{theorem}\label{thm:periodII} For $0