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∑ k个 = 0 ∞ ( 1 4 k个 + 1 − 1 4 k个 + 三 ) = π 4 {\displaystyle\sum_{k=0}^{\infty}\left({\frac{1}{4k+1}}-{\frac{1}{4k+3}}\right)={\ frac{\pi}{4}}}}
将两边乘以4,
∑ k个 = 0 ∞ ( 4 4 k个 + 1 − 4 4 k个 + 三 ) = π {\displaystyle\sum_{k=0}^{\infty}\left({\frac{4}{4k+1}}-{\frac{4}{4k+3}}\right)=\pi} (1)
∑ k个 = 0 ∞ ( 1 4 k个 + 1 − 1 4 k个 + 5 ) = 1 {\displaystyle\sum_{k=0}^{\infty}\left({\frac{1}{4k+1}}-{\frac{1}{4k+5}}\right)=1}
将两边乘以c,
∑ k个 = 0 ∞ ( c(c) 4 k个 + 1 − c(c) 4 k个 + 5 ) = c(c) {\displaystyle\sum_{k=0}^{\infty}\left({\frac{c}{4k+1}}-{\frac{c}{4k+5}}\right)=c} (2)
从方程式(1)中减去方程式(2),
∑ k个 = 0 ∞ ( 4 − c(c) 4 k个 + 1 − 4 4 k个 + 三 + c(c) 4 k个 + 5 ) = π − c(c) {\displaystyle\sum_{k=0}^{\infty}\left({\frac{4-c}{4k+1}}-{\frac{4}{4k+3}}+{\frac:c}{4 k+5}}}\right)=\pi-c}