$\sum _{k=0}^{\infty }\left({\frac {1}{4k+1}}-{\frac {1}{4k+3}}\right)={\frac {\pi }{4}}$

Multiplying both sides by 4,

$\sum _{k=0}^{\infty }\left({\frac {4}{4k+1}}-{\frac {4}{4k+3}}\right)=\pi$ (1)

$\sum _{k=0}^{\infty }\left({\frac {1}{4k+1}}-{\frac {1}{4k+5}}\right)=1$

Multiplying both sides by c,

$\sum _{k=0}^{\infty }\left({\frac {c}{4k+1}}-{\frac {c}{4k+5}}\right)=c$ (2)

Subtracting equation (2) from equation (1),

$\sum _{k=0}^{\infty }\left({\frac {4-c}{4k+1}}-{\frac {4}{4k+3}}+{\frac {c}{4k+5}}\right)=\pi -c$