Yet another π formula

[Gerson W. Barbosa]

(01-06-2021 01:32 AM)Albert Chan Wrote:  

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

Proof:

atan(x) = x - x^3/3 + x^5/5 - x^7/7 + ...
atan(1) = pi/4 = 1 - 1/3 + 1/5 - 1/7 + ...

2*I = 2/(1*3) - 2/(3*5) + 2/(5*7) - 2/(7*9) + ...
      = (1-1/3) - (1/3-1/5) + (1/5-1/7) - (1/7-1/9) + ...
      = (1-1/3+1/5-1/7+...) + (-1/3+1/5-1/7+...)
      =           pi/4               +      (pi/4 - 1)

→ I = pi/4 - 1/2

Great! The continued fraction part is left unproved, however. But that would not be easy, I presume.