Yet another π formula

[Albert Chan]

From previous post, if we ignore S correction, with goal of c1 * c2 = pi, we get this.

\(\displaystyle \frac{\pi}{2} = \sum _{k=0}^∞ \frac {k!}{(2k+1)!!}\)

Formula approximately doubled its accuracy after each sum.
This explained why previous post, with N=40, c1*c2 get about 40 bits accuracy of pi.

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To speed up convergence, we can "double" the last term.

pi/2 = 1 + 1/3 + 2/15 + 6/105 + 24/945 + ...
       = (2-1) + (2/3-1/3) + (4/15-2/15) + (12/105-6/105) + (48/945-24/945) + ...
       = 2 - (1-2/3) - (1/3-4/15) - (2/15-12/105) - (6/105-48/945) - ...

k!/(2k+1)!! - 2*(k+1)!/(2k+3)!! = k!/(2k+3)!!

Again, we have a pattern here

Code:

pi/2 = 0 + sum(k!/(2k+1)!!, k=0..inf)
pi/2 = 2 - sum(k!/(2k+3)!!, k=0..inf)             -- 0 + 2/1 = 2
pi/2 = 4/3 + 3!! * sum(k!/(2k+5)!!, k=0..inf)     -- 2 - 2/3 = 4/3
pi/2 = 26/15 - 5!! * sum(k!/(2k+7)!!, k=0..inf)   -- 4/3 + 2/5 = 26/15
pi/2 = 152/105 + 7!! * sum(k!/(2k+9)!!, k=0..inf) -- 26/15 - 2/7 = 152/105
...

Ignore the sum part, we goes full circle

pi/2 = 2/1 - 2/3 + 2/5 - 2/7 + ... = 2 * atan(1)