Alternating Near Harmonic Series?

Thanks, Abhishek Ghosh for the nice solution. That would mean it will work with $\tan^{-1} {x}$ too. Which is also more familiar to most people.

The Maclaurin series of $\tan^{-1} {x}$ is as follows:

$\quad \quad \tan^{-1} {x} = x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 +...$

We know that:

$\quad \quad \tan^{-1} {1} = \frac {\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} +...$

$\quad \quad \Rightarrow \pi = 4 \times \left( 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} +... \right)$

By approximation. The terms of the series are decreasing so the answer is somewhere between three and four. The gravitational constant is about ten, euler-mascheroni constant is about a half, pi is between three and four and the planck constant is about zero. So it has to be pi....

Just for those who aren't much familiar with taylor and mclaurin's theorem,they can follow this approach Let f(x)=1/1+x^2=(1+x^2)^(-1),,now expand using binomial theorem for any rational index to obtain f(x)=1-x^2+x^4-x^6+....... Now integrate both sides and put x=1,to obtain pi/4...which when multiplied by 4 gives us pi..:)

Python 2.7:

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from time import time
from fractions import Fraction as frac
total = 0
denominator = 1
sign = 1
end = time() + 30
# stick your pi in the oven for 30 seconds
while time() < end:
    total += frac(1,denominator*sign)
    denominator += 2
    sign *= -1
print "Answer:", 4 * float(total)
# ding!

I love this theorem .