Pi

$$\pi =4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...)$$

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 $\frac{d}{\mathrm{dx}}{\tan }^{-1}x=\frac{1}{1+x^2}$, 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:

$${\tan }^{-1}(1)=4{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}\frac{1}{239}$$