The Gregory-Leibniz Series
\[\pi = 4{(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ...)}\]
Proof: Start with the Taylor series:
\[\frac{1}{1-y}
=
1 + y + y^2 + ...\]
Apply the variable substitution \(y = -x^2\) to get
\[\frac{1}{1+x^2}
=
1 - x^2 + x^4 - x^6 + ...\]
Now since \(\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}\), by integrating, we find that the Taylor expansion of \(\tan^{-1} x\) is
\[\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - ...\]
and the formula is obtained by substituting \(x = 1\).
Variations
The Gregory-Leibniz Series converges very slowly. One way to improve it is to use
\[\tan^{-1}\frac{1}{\sqrt{3}} = \pi/6
= \frac{1}{\sqrt{3}}{(1 - \frac{1}{3\cdot 3}
+ \frac {1}{5\cdot 3^2}
- \frac {1}{7\cdot 3^3}
+ ...)}\]
Even better is
\[\tan^{-1}1 = \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}
= \frac{1}{2}{(1 - \frac{1}{3\cdot 2^2} + \frac{1}{5\cdot 2^4} - ...)}
+ \frac{1}{3}{(1 - \frac{1}{3\cdot 3^2} + \frac{1}{5\cdot 3^4} - ...)}\]
Another way that is handy for decimal digits is:
\[\tan^{-1}(1) = 4 \tan^{-1}\frac{1}{5} - \tan^{-1}\frac{1}{239}\]