1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯ = π 4 1 - {1 \over 3} + {1 \over 5} - {1 \over 7} + {1 \over 9} - \cdots = {\pi \over 4}
∑ n=0 ∞ (−1 )n 2n+1 = π 4 \sum_{n=0}^\infty {(-1)^n \over 2n + 1} = {\pi \over 4}
1 π = 2 2 99 2 ∑ n=0 ∞ (4n)!(1103+26390n) ( 4n 99nn!)4 {1 \over \pi} = {2 \sqrt{2} \over 99^2}\sum_{n=0}^\infty {(4n)!(1103 + 26390n) \over (4^n99^nn!)^4}