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∑ k = 0 ∞ ( 1 4 k + 1 − 1 4 k + 3 ) = π 4 {\displaystyle \sum _{k=0}^{\infty }\left({\frac {1}{4k+1}}-{\frac {1}{4k+3}}\right)={\frac {\pi }{4}}}
Multiplying both sides by 4,
∑ k = 0 ∞ ( 4 4 k + 1 − 4 4 k + 3 ) = π {\displaystyle \sum _{k=0}^{\infty }\left({\frac {4}{4k+1}}-{\frac {4}{4k+3}}\right)=\pi } (1)
∑ k = 0 ∞ ( 1 4 k + 1 − 1 4 k + 5 ) = 1 {\displaystyle \sum _{k=0}^{\infty }\left({\frac {1}{4k+1}}-{\frac {1}{4k+5}}\right)=1}
Multiplying both sides by c,
∑ k = 0 ∞ ( c 4 k + 1 − c 4 k + 5 ) = c {\displaystyle \sum _{k=0}^{\infty }\left({\frac {c}{4k+1}}-{\frac {c}{4k+5}}\right)=c} (2)
Subtracting equation (2) from equation (1),
∑ k = 0 ∞ ( 4 − c 4 k + 1 − 4 4 k + 3 + c 4 k + 5 ) = π − c {\displaystyle \sum _{k=0}^{\infty }\left({\frac {4-c}{4k+1}}-{\frac {4}{4k+3}}+{\frac {c}{4k+5}}\right)=\pi -c}
(1)
(2)
