Let $\eta(s) = \sum_{m=1}^\infty \frac{(-1)^{m-1}}{m^s} = (1 - 2^{1-s})\zeta(s)$ denote the Dirichlet eta function, and $\beta(s) = \sum_{m=0}^\infty \frac{(-1)^m}{(2m+1)^s}$ denote the Dirichlet beta function.
We define the relative parity ratio $R(n)$:
$$R(n) = \frac{\beta(n)}{\eta(n)}$$The normalized parity remainder $k(n)$ isolates the scaled excess above unity:
$$k(n) = 2^n \left( R(n) - 1 \right)$$Applying the Cayley transform to the parity remainder $k(n)$ yields:
$$\mathcal{C}(n) = \frac{1 - k(n)}{1 + k(n)} = \frac{(1 + 2^n) - 2^n R(n)}{(1 - 2^n) + 2^n R(n)}$$The complementary orthogonal component $\mathcal{W}(n)$, satisfying the identity $\mathcal{C}(n)^2 + \mathcal{W}(n)^2 = 1$, is given by:
$$\mathcal{W}(n) = \frac{2\sqrt{k(n)}}{1 + k(n)} = \frac{2\sqrt{2^n (R(n) - 1)}}{1 + 2^n (R(n) - 1)}$$| $n$ | $\beta(n)$ | $\eta(n)$ | $R(n)$ | $k(n)$ | $\mathcal{C}(n)$ | $\mathcal{W}(n)$ | $\mathcal{C}^2 + \mathcal{W}^2$ |
|---|---|---|---|---|---|---|---|
| 2 | 0.9159655942 | 0.8224670334 | 1.1136807005 | 0.4547227806 | 0.3748317808 | 0.9270921447 | 1.0000000000 |
| 3 | 0.9689461463 | 0.9015426774 | 1.0747646900 | 0.5981175198 | 0.2514725946 | 0.9678643813 | 1.0000000000 |
| 4 | 0.9889445517 | 0.9470328295 | 1.0442558283 | 0.7080932525 | 0.1708962601 | 0.9852890280 | 1.0000000000 |
| 5 | 0.9961578281 | 0.9721197704 | 1.0246718399 | 0.7894988771 | 0.1176313702 | 0.9930573981 | 1.0000000000 |
| 6 | 0.9986852222 | 0.9855510913 | 1.0132684659 | 0.8491823798 | 0.0815590924 | 0.9966687352 | 1.0000000000 |
| 7 | 0.9995545079 | 0.9926037456 | 1.0069723521 | 0.8926955091 | 0.0566939921 | 0.9983915124 | 1.0000000000 |
| 8 | 0.9998499902 | 0.9962491780 | 1.0036143537 | 0.9242635955 | 0.0393586524 | 0.9992251470 | 1.0000000000 |
| 9 | 0.9999496842 | 0.9981090628 | 1.0018441011 | 0.9470366657 | 0.0272020211 | 0.9996299818 | 1.0000000000 |
| 10 | 0.9999831640 | 0.9990500164 | 1.0009340320 | 0.9633887225 | 0.0186469812 | 0.9998261317 | 1.0000000000 |
As $n \to \infty$, $R(n) \to 1 + 2^{-n} + \mathcal{O}(3^{-n})$, driving $k(n) \to 1$ from below. In this limit, $\mathcal{C}(n) \to 0$ and $\mathcal{W}(n) \to 1$.
At the pole boundary $s = 1$, taking the limit yields:
$$k(1) = 2\left(\frac{\pi/4}{\ln 2} - 1\right) \approx 0.266180$$corresponding to $\mathcal{C}(1) \approx 0.579554$ and $\mathcal{W}(1) \approx 0.814934$.
The discrete reflection point $k = -1$ maps to $\mathcal{C} \to \infty$ and $\mathcal{W}$ entering the imaginary domain, representing the orthogonal boundary where the sum coordinate vanishes.