Find the challenging infinite sum

Consider one stage in this construction process, where one right-angled triangle has angle $\theta$ and side $x$ , and the next right-angled triangle has angle $\phi$ and corresponding side $y$ . Then $\begin{aligned} \frac{\sin(\theta+\phi)}{x} & = \; \frac{\sin\phi}{t x \sec\theta} \\ t\sec\theta \sin(\theta + \phi) & = \; \sin\phi \\ t\tan\theta \cos\phi + t\sin\phi & = \; \sin\phi \\ \tan\phi & = \; \frac{t}{1-t}\tan\theta \end{aligned}$ Moreover, we note that $y \; = \; x \tan\theta \cos\phi$

Thus, if $\theta_n$ (for $n \ge 0$ ) is the corresponding angle of the $n$ th right-angled triangle, then $\tan\theta_{n+1} \; = \; \frac{t}{1-t} \tan\theta_n$ and hence $\tan\theta_n \; = \; \left(\frac{t}{1-t}\right)^n \tan\theta \hspace{2cm} n \ge 0$ Moreover, if $x_n$ is the corresponding side of the $n$ th triangle, then $x_{n+1} \; = \; x_n \tan\theta_n \cos\theta_{n+1}$ and the area of the $n$ th triangle is $S_n \; = \; \tfrac12x_n^2\tan\theta_n$ so that $\begin{aligned} S_{n+1} & = \; \tfrac12x_{n+1}^2 \tan\theta_{n+1} \; = \; \tfrac12x_n^2 \tan^2\theta_n \cos^2\theta_{n+1}\tan\theta_{n+1} \; = \; S_n \tan\theta_n \tan\theta_{n+1}\cos^2\theta_{n+1} \\ & = \; S_n \left(\frac{t}{1-t}\right)^{2n+1} \tan^2\theta \cos^2\theta_{n+1} \end{aligned}$ and so $\begin{aligned} \left(\frac{1-t}{t}\right)^{(n+1)^2}S_{n+1} &= \; \left(\frac{1-t}{t}\right)^{n^2}S_n \tan^2\theta \cos^2\theta_{n+1}\\ \left(\frac{1-t}{t}\right)^{(n+1)^2}\frac{S_{n+1}}{\tan^{2(n+1)}\theta} &= \; \left(\frac{1-t}{t}\right)^{n^2}\frac{S_n}{\tan^{2n}\theta} \cos^2\theta_{n+1} \end{aligned}$ so we deduce that $S_n \; = \; S_0 \left(\frac{t}{1-t}\right)^{n^2} \tan^{2n}\theta \prod_{j=1}^n \cos^2\theta_j \; = \; \frac12 \left(\frac{t}{1-t}\right)^{n^2} \tan^{2n+1}\theta \prod_{j=1}^n \cos^2\theta_j$ Thus it follows that $\lim_{\theta \to 0+} \frac{S_n}{\theta^{2n+1}} \; =\; \frac12\left(\frac{t}{1-t}\right)^{n^2}$ so we deduce that $a_n \; = \; 2n+1 \hspace{1cm} b_n \; = \; \frac12\left(\frac{t}{1-t}\right)^{n^2} \hspace{2cm} n \ge 0$ Since $b_0 = \tfrac12$ and we require the sum of the $b_n$ to be less than $\tfrac12$ , we must be ignoring the original triangle $ABC$ in these calculations, and we need to find $0 < t <1$ such that $1 \; = \; 2\sum_{n=1}^\infty b_n \; = \; \sum_{n=1}^\infty q^{n^2} \; =\; \frac12\left( \vartheta_3(0,q) - 1\right)$ where $q = \tfrac{t}{1-t}$ . Solving this equation numerically gives $q \; = \; 0.705346681379806989636379706394... \hspace{2cm} t \; = \; 0.413608968241640136350617724053$ and hence $X \; =\; \sum_{n=1}^{\infty}a_nb_n \; = \; \frac12\sum_{n=1}^{\infty}(2n+1)q^{n^2} \; = \; 1.84587508767744720084799289178$ and hence $\lfloor 10^{10}X \rfloor \; = \; \boxed{18458750876}$ Although I am not immediately aware of a closed form for $X$ , it is easy to find $X$ to the required degree of accuracy. Since $q < \tfrac34$ , we can show that $(2n+1)q^n \le 7q^3 < 3 \hspace{2cm} n \in \mathbb{N}$ and hence $\sum_{n=N+1}^\infty (2n+1)q^{n^2} \; \le \; 3\sum_{n=N+1}^\infty q^{n(n-1)} \; \le \; 3q^N \sum_{n=N+1}^\infty q^{(n-1)^2} \; \le \; 3q^N \sum_{n=N^2}^\infty q^n \; = \; 12\big(\tfrac34\big)^{N(N+1)}$ Thus $\sum_{n=N+1}^\infty (2n+1)q^{n^2} \; < \; 5 \times 10^{-11}$ provided that $12\big(\tfrac34\big)^{N(N+1)} < 5\times10^{-11}$ or $N(N+1) > \frac{11+\log_{10}2.4}{\log_{10}\frac43} = 91.09$ , and hence provided that $N \ge 10$ . Thus we can get $10$ decimal place accuracy for $X$ by calculating $\sum_{n=1}^{10}(2n+1)q^{n^2}$