Yet another π formula

[Gerson W. Barbosa]

(01-05-2021 10:50 PM)Albert Chan Wrote:  

(01-04-2021 08:41 PM)Gerson W. Barbosa Wrote:  The alternate sum of the factors of the Wallis product tends to \(\pi\)/4 - 1/2 as \(n\) tends to infinity:

\(\lim_{n\rightarrow \infty } \left [ \frac{1}{1\cdot 3}-\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}-\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}-\frac{1}{11\cdot 13}+\frac{1}{13\cdot 15}-\frac{1}{15 \cdot 17}\pm \cdots +\frac{(-1)^{n-1}}{4n^{2}-1}\right ]=\frac{\pi }{4}-\frac{1}{2}\)

This converges to the constant significantly faster than the Wallis Product.
For 12 correct digits of \(\pi\) it requires \(\sqrt{10^{12}}\) terms instead of the \(10^{12}\) terms required by Wallis.

Where does this limit comes from ?

Sum(n=1, inf, (-1)^(n-1)/(4n^2-1)) (WolframAlpha)