tanleft(frac{pi}{4}right) &=& 1 
frac{pi}{4} &=& ta...
&=& 1 - frac{1}{3} + frac{1}{5} - frac{1}{7} + cdots

(Proofs of any of the individual steps are available upon request, should you find yourself thinking that I’m pulling a 1=0 trick.) So then

begin{displaymath}pi = 4left(1 - frac{1}{3} + frac{1}{5} - frac{1}{7} + cdotsright).end{displaymath}

This converges very slowly, though, because for every two steps forward you take
a step back. (More precisely: for every 1 step forward, you take $(4n+1)/(4n+3)$ steps back.) You can make it converge faster by combining the forward step and the smaller backward step into a single, smaller, forward step:

frac{pi}{4} = sum_{n=0}^{+infty} frac{(-1)^n}{2n+1} &=& ...
...n+3} right) 
&=& sum_{n=0}^{+infty} frac{2}{(4n+1)(4n+3)}


begin{displaymath}pi = 8left( frac{1}{3} + frac{1}{35} + frac{1}{99} + cdotsright)end{displaymath}