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 hold exactly. Passing to the limit yields
3-\mathbf{K}_{n\ge0}\frac{b_n}{a_{n+1}}=\frac{16}{\pi^2-4}.
Thus:
3 – \cfrac{3}{13 – \cfrac{48}{29 – \cfrac{225}{51 – \cfrac{672}{79 – \ddots}}}}
= \frac{16}{\pi^2-4}.
