Author: Lezhe

X := 1+\mathbf{K}_{n\ge1}\frac{(2n-1)^2}{2}=\frac{4}{\pi}. Apply Euler’s even/odd recombination to finite truncations of \(X\). Obtain an equivalent continued fraction. Apply the Bauer–Muir transform with parameter r_n=n(n+3). After simplification the coefficients are a_n=3n^2+7n+3,\qquad b_n=(n+1)^2(n+3)(2n+1). The resulting continued fraction is T(X)=3-\mathbf{K}_{n\ge0}\frac{b_n}{a_{n+1}}. Define T(t)=\frac{4t^2}{4-t^2}. Since \(X=\tfrac{4}{\pi}\), T(X)=\frac{4(4/\pi)^2}{4-(4/\pi)^2}=\frac{16}{\pi^2-4}. Finite truncation equalities …

where for \(n\ge1\), \[ a_n=3n^2+7n+4,\quad b_n=-2n^2(n+1)^2,\quad a_0=4. \] Proof: 1. Define convergents: \[ \frac{P_n}{Q_n}=a_0+\cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cdots+\cfrac{b_n}{a_n}}}. \] Recurrence relations: \[ P_{-1}=1,\;P_0=a_0=4,\quad Q_{-1}=0,\;Q_0=1, \] \[ P_n=a_nP_{n-1}+b_nP_{n-2},\quad Q_n=a_nQ_{n-1}+b_nQ_{n-2}. \] 2. By induction one finds \[ P_n=2^{\,n+1}(n+1)!(n+2)!. \] 3. Define \[ y_n=\frac{Q_n}{P_n},\quad \frac{P_n}{Q_n}=\frac{1}{y_n}. \] From recurrence we obtain \[ z_n:=y_n-y_{n-1}=-\frac{b_nP_{n-2}}{P_n}(y_{n-1}-y_{n-2}). …