∑ n = 0 ∞ 2 − n = 1 + 1 2 + 1 4 + 1 8 + 1 16 + ⋯ = 2 {\displaystyle \sum _{n=0}^{\infty }2^{-n}=1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots =2}
∑ n = 0 ∞ 2 n = 1 + 2 + 4 + 8 + 16 + ⋯ = − 1 {\displaystyle \sum _{n=0}^{\infty }2^{n}=1+2+4+8+16+\cdots =-1}
∑ n = 0 ∞ ( − 2 ) − n = 1 − 1 2 + 1 4 − 1 8 + 1 16 − ⋯ = 2 3 {\displaystyle \sum _{n=0}^{\infty }(-2)^{-n}=1-{\frac {1}{2}}+{\frac {1}{4}}-{\frac {1}{8}}+{\frac {1}{16}}-\cdots ={\frac {2}{3}}}
∑ n = 0 ∞ ( − 2 ) n = 1 − 2 + 4 − 8 + 16 − ⋯ = 1 3 {\displaystyle \sum _{n=0}^{\infty }(-2)^{n}=1-2+4-8+16-\cdots ={\frac {1}{3}}}
∑ n = 0 ∞ 2 − 2 n = 1 + 1 4 + 1 16 + 1 64 + 1 256 + ⋯ = 4 3 {\displaystyle \sum _{n=0}^{\infty }2^{-2n}=1+{\frac {1}{4}}+{\frac {1}{16}}+{\frac {1}{64}}+{\frac {1}{256}}+\cdots ={\frac {4}{3}}}
∑ n = 0 ∞ 2 2 n = 1 + 4 + 16 + 64 + 256 + ⋯ = − 1 3 {\displaystyle \sum _{n=0}^{\infty }2^{2n}=1+4+{16}+{64}+{256}+\cdots =-{\frac {1}{3}}}
∑ n = 0 ∞ ( 2 i ) − 2 n = 1 − 1 4 + 1 16 − 1 64 + 1 256 − ⋯ = 4 5 {\displaystyle \sum _{n=0}^{\infty }(2i)^{-2n}=1-{\frac {1}{4}}+{\frac {1}{16}}-{\frac {1}{64}}+{\frac {1}{256}}-\cdots ={\frac {4}{5}}}
∑ n = 0 ∞ ( 2 i ) 2 n = 1 − 4 + 16 − 64 + 256 − ⋯ = 1 5 {\displaystyle \sum _{n=0}^{\infty }(2i)^{2n}=1-4+{16}-{64}+{256}-\cdots ={\frac {1}{5}}}
∑ n = 0 ∞ 1 = 1 + 1 + 1 + 1 + 1 + ⋯ = − 1 2 {\displaystyle \sum _{n=0}^{\infty }1=1+1+1+1+1+\cdots =-{\frac {1}{2}}}
∑ n = 0 ∞ ( − 1 ) n = 1 − 1 + 1 − 1 + 1 − ⋯ = 1 2 {\displaystyle \sum _{n=0}^{\infty }(-1)^{n}=1-1+1-1+1-\cdots ={\frac {1}{2}}}
∑ n = 0 ∞ − 1 = − 1 − 1 − 1 − 1 − 1 − ⋯ = 1 2 {\displaystyle \sum _{n=0}^{\infty }-1=-1-1-1-1-1-\cdots ={\frac {1}{2}}}
∑ n = 0 ∞ ( − 1 ) n + 1 = − 1 + 1 − 1 + 1 − 1 + ⋯ = − 1 2 {\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}=-1+1-1+1-1+\cdots =-{\frac {1}{2}}}
∑ n = 0 ∞ n = 0 + 1 + 2 + 3 + 4 + ⋯ = − 1 12 {\displaystyle \sum _{n=0}^{\infty }n=0+1+2+3+4+\dotsb =-{\frac {1}{12}}}
∑ n = 0 ∞ ( − 1 ) n n = 0 − 1 + 2 − 3 + 4 − ⋯ = − 1 4 {\displaystyle \sum _{n=0}^{\infty }(-1)^{n}n=0-1+2-3+4-\dotsb =-{\frac {1}{4}}}
∑ n = 0 ∞ − n = 0 − 1 − 2 − 3 − 4 − ⋯ = 1 12 {\displaystyle \sum _{n=0}^{\infty }-n=0-1-2-3-4-\dotsb ={\frac {1}{12}}}
∑ n = 0 ∞ ( − 1 ) n + 1 n = 0 + 1 − 2 + 3 − 4 + ⋯ = 1 4 {\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}n=0+1-2+3-4+\dotsb ={\frac {1}{4}}}
∑ n = 0 ∞ 2 n = 0 + 2 + 4 + 6 + 8 + ⋯ = − 1 6 {\displaystyle \sum _{n=0}^{\infty }2n=0+2+4+6+8+\dotsb =-{\frac {1}{6}}}
∑ n = 0 ∞ ( 2 n + 1 ) = 1 + 3 + 5 + 7 + 9 + ⋯ = − 2 3 {\displaystyle \sum _{n=0}^{\infty }(2n+1)=1+3+5+7+9+\dotsb =-{\frac {2}{3}}}
∑ n = 1 ∞ 1 n = 1 + 1 2 + 1 3 + 1 4 + 1 5 + ⋯ = ∞ {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots =\infty }
∑ n = 1 ∞ ( − 1 ) n + 1 n = 1 − 1 2 + 1 3 − 1 4 + 1 5 − ⋯ = ln 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-\cdots =\ln 2}
∑ n = 0 ∞ 1 2 n + 1 = 1 + 1 3 + 1 5 + 1 7 + 1 9 + ⋯ = ∞ {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{2n+1}}=1+{\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{9}}+\cdots =\infty }
∑ n = 0 ∞ ( − 1 ) n 2 n + 1 = 1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯ = π 4 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots ={\frac {\pi }{4}}}
∑ n = 0 ∞ 1 ( 2 n + 1 ) ! = 1 + 1 3 ! + 1 5 ! + 1 7 ! + 1 9 ! + ⋯ = e 2 − 1 2 e {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(2n+1)!}}=1+{\frac {1}{3!}}+{\frac {1}{5!}}+{\frac {1}{7!}}+{\frac {1}{9!}}+\cdots ={\frac {e}{2}}-{\frac {1}{2e}}}
∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! = 1 − 1 3 ! + 1 5 ! − 1 7 ! + 1 9 ! − ⋯ = sin 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}=1-{\frac {1}{3!}}+{\frac {1}{5!}}-{\frac {1}{7!}}+{\frac {1}{9!}}-\cdots =\sin 1}
∑ n = 0 ∞ 1 ( 2 n ) ! = 1 + 1 2 ! + 1 4 ! + 1 6 ! + 1 8 ! + ⋯ = e 2 + 1 2 e {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(2n)!}}=1+{\frac {1}{2!}}+{\frac {1}{4!}}+{\frac {1}{6!}}+{\frac {1}{8!}}+\cdots ={\frac {e}{2}}+{\frac {1}{2e}}}
∑ n = 0 ∞ ( − 1 ) n ( 2 n ) ! = 1 − 1 2 ! + 1 4 ! − 1 6 ! + 1 8 ! − ⋯ = cos 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}=1-{\frac {1}{2!}}+{\frac {1}{4!}}-{\frac {1}{6!}}+{\frac {1}{8!}}-\cdots =\cos 1}
∑ n = 0 ∞ 1 n ! = 1 + 1 + 1 2 + 1 6 + 1 24 + 1 120 + ⋯ = e {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n!}}=1+1+{\frac {1}{2}}+{\frac {1}{6}}+{\frac {1}{24}}+{\frac {1}{120}}+\cdots =e}
∑ n = 0 ∞ ( − 1 ) n n ! = 1 − 1 + 1 2 − 1 6 + 1 24 − 1 120 + ⋯ = 1 e {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}=1-1+{\frac {1}{2}}-{\frac {1}{6}}+{\frac {1}{24}}-{\frac {1}{120}}+\cdots ={\frac {1}{e}}}
∑ n = 0 ∞ ( − 1 ) n π 2 n + 1 ( 2 n + 1 ) ! = 1 − π 3 3 ! + π 5 5 ! − π 7 7 ! + π 9 9 ! − ⋯ = − 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}\pi ^{2n+1}}{(2n+1)!}}=1-{\frac {\pi ^{3}}{3!}}+{\frac {\pi ^{5}}{5!}}-{\frac {\pi ^{7}}{7!}}+{\frac {\pi ^{9}}{9!}}-\cdots =-1}
∑ n = 0 ∞ ( − 1 ) n π 2 n ( 2 n ) ! = 1 − π 2 2 ! + π 4 4 ! − π 6 6 ! + π 8 8 ! − ⋯ = 0 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}\pi ^{2n}}{(2n)!}}=1-{\frac {\pi ^{2}}{2!}}+{\frac {\pi ^{4}}{4!}}-{\frac {\pi ^{6}}{6!}}+{\frac {\pi ^{8}}{8!}}-\cdots =0}
∑ n = 0 ∞ ( − 1 ) n π 2 n + 1 ( 2 n + 1 ) ! 2 2 n + 1 = 1 − π 3 3 ! 2 3 + π 5 5 ! 2 5 − π 7 7 ! 2 7 + π 9 9 ! 2 9 − ⋯ = 0 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}\pi ^{2n+1}}{(2n+1)!2^{2n+1}}}=1-{\frac {\pi ^{3}}{3!2^{3}}}+{\frac {\pi ^{5}}{5!2^{5}}}-{\frac {\pi ^{7}}{7!2^{7}}}+{\frac {\pi ^{9}}{9!2^{9}}}-\cdots =0}
∑ n = 0 ∞ ( − 1 ) n π 2 n ( 2 n ) ! 2 2 n = 1 − π 2 2 ! 2 2 + π 4 4 ! 2 4 − π 6 6 ! 2 6 + π 8 8 ! 2 8 − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}\pi ^{2n}}{(2n)!2^{2n}}}=1-{\frac {\pi ^{2}}{2!2^{2}}}+{\frac {\pi ^{4}}{4!2^{4}}}-{\frac {\pi ^{6}}{6!2^{6}}}+{\frac {\pi ^{8}}{8!2^{8}}}-\cdots =1}
∑ n = 0 ∞ ( − 1 ) n π 2 n + 1 ( 2 n + 1 ) ! 2 4 n + 2 = 1 − π 3 3 ! 2 6 + π 5 5 ! 2 10 − π 7 7 ! 2 14 + π 9 9 ! 2 18 − ⋯ = 1 2 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}\pi ^{2n+1}}{(2n+1)!2^{4n+2}}}=1-{\frac {\pi ^{3}}{3!2^{6}}}+{\frac {\pi ^{5}}{5!2^{10}}}-{\frac {\pi ^{7}}{7!2^{14}}}+{\frac {\pi ^{9}}{9!2^{18}}}-\cdots ={\frac {1}{\sqrt {2}}}}
∑ n = 0 ∞ ( − 1 ) n π 2 n ( 2 n ) ! 2 4 n = 1 − π 2 2 ! 2 4 + π 4 4 ! 2 8 − π 6 6 ! 2 12 + π 8 8 ! 2 16 − ⋯ = 1 2 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}\pi ^{2n}}{(2n)!2^{4n}}}=1-{\frac {\pi ^{2}}{2!2^{4}}}+{\frac {\pi ^{4}}{4!2^{8}}}-{\frac {\pi ^{6}}{6!2^{12}}}+{\frac {\pi ^{8}}{8!2^{16}}}-\cdots ={\frac {1}{\sqrt {2}}}}
∑ n = 0 ∞ ( i π ) n n ! = 1 + i π − π 2 2 − i π 3 6 + π 4 24 + i π 5 120 − ⋯ = − 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(i\pi )^{n}}{n!}}=1+i\pi -{\frac {\pi ^{2}}{2}}-{\frac {i\pi ^{3}}{6}}+{\frac {\pi ^{4}}{24}}+{\frac {i\pi ^{5}}{120}}-\cdots =-1}
∑ n = 0 ∞ i n = 1 + i − 1 − i + 1 + ⋯ = 1 + i 2 {\displaystyle \sum _{n=0}^{\infty }i^{n}=1+i-1-i+1+\cdots ={\frac {1+i}{2}}}
∑ n = 0 ∞ i − n = 1 − i − 1 + i + 1 − ⋯ = 1 − i 2 {\displaystyle \sum _{n=0}^{\infty }i^{-n}=1-i-1+i+1-\cdots ={\frac {1-i}{2}}}
∑ n = 0 ∞ ( − i ) n = 1 − i − 1 + i + 1 − ⋯ = 1 − i 2 {\displaystyle \sum _{n=0}^{\infty }(-i)^{n}=1-i-1+i+1-\cdots ={\frac {1-i}{2}}}
∑ n = 0 ∞ ( − i ) − n = 1 + i − 1 − i + 1 + ⋯ = 1 + i 2 {\displaystyle \sum _{n=0}^{\infty }(-i)^{-n}=1+i-1-i+1+\cdots ={\frac {1+i}{2}}}
∑ n = 0 ∞ 2 − n − 1 = 1 2 + 1 4 + 1 8 + 1 16 + 1 32 + ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }2^{-n-1}={\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+{\frac {1}{32}}+\cdots =1}
∑ n = 0 ∞ − 2 n = − 1 − 2 − 4 − 8 − 16 − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }-2^{n}=-1-2-4-8-16-\cdots =1}
∑ n = 0 ∞ 3 2 ( − 2 ) − n = 3 2 − 3 4 + 3 8 − 3 16 + 3 32 − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {3}{2}}(-2)^{-n}={\frac {3}{2}}-{\frac {3}{4}}+{\frac {3}{8}}-{\frac {3}{16}}+{\frac {3}{32}}-\cdots =1}
∑ n = 0 ∞ 3 ( − 2 ) n = 3 − 6 + 12 − 24 + 48 − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }3(-2)^{n}=3-6+12-24+48-\cdots =1}
∑ n = 0 ∞ − 2 = − 2 − 2 − 2 − 2 − 2 − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }-2=-2-2-2-2-2-\cdots =1}
∑ n = 0 ∞ ( − 1 ) n 2 = 2 − 2 + 2 − 2 + 2 − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }(-1)^{n}2=2-2+2-2+2-\cdots =1}
∑ n = 0 ∞ − 12 n = 0 − 12 − 24 − 36 − 48 − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }-12n=0-12-24-36-48-\dotsb =1}
∑ n = 0 ∞ − ( − 1 ) n 4 n = 0 + 4 − 8 + 12 − 16 + ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }-(-1)^{n}4n=0+4-8+12-16+\dotsb =1}
∑ n = 1 ∞ ( − 1 ) n + 1 n ln 2 = 1 ln 2 − 1 2 ln 2 + 1 3 ln 2 − 1 4 ln 2 + 1 5 ln 2 − ⋯ = 1 {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n\ln 2}}={\frac {1}{\ln 2}}-{\frac {1}{2\ln 2}}+{\frac {1}{3\ln 2}}-{\frac {1}{4\ln 2}}+{\frac {1}{5\ln 2}}-\cdots =1}
∑ n = 0 ∞ 4 π ( − 1 ) n 2 n + 1 = 4 π − 4 3 π + 4 5 π − 4 7 π + 4 9 π − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {4}{\pi }}{\frac {(-1)^{n}}{2n+1}}={\frac {4}{\pi }}-{\frac {4}{3\pi }}+{\frac {4}{5\pi }}-{\frac {4}{7\pi }}+{\frac {4}{9\pi }}-\cdots =1}
∑ n = 0 ∞ 2 e e 2 − 1 1 ( 2 n + 1 ) ! = 2 e e 2 − 1 + 2 e 3 ! ( e 2 − 1 ) + 2 e 5 ! ( e 2 − 1 ) + 2 e 7 ! ( e 2 − 1 ) + 2 e 9 ! ( e 2 − 1 ) + ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {2e}{e^{2}-1}}{\frac {1}{(2n+1)!}}={\frac {2e}{e^{2}-1}}+{\frac {2e}{3!(e^{2}-1)}}+{\frac {2e}{5!(e^{2}-1)}}+{\frac {2e}{7!(e^{2}-1)}}+{\frac {2e}{9!(e^{2}-1)}}+\cdots =1}
∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! sin 1 = 1 sin 1 − 1 3 ! sin 1 + 1 5 ! sin 1 − 1 7 ! sin 1 + 1 9 ! sin 1 − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!\sin 1}}={\frac {1}{\sin 1}}-{\frac {1}{3!\sin 1}}+{\frac {1}{5!\sin 1}}-{\frac {1}{7!\sin 1}}+{\frac {1}{9!\sin 1}}-\cdots =1}
∑ n = 0 ∞ 2 e e 2 + 1 1 ( 2 n ) ! = 2 e e 2 + 1 + 2 e 2 ! ( e 2 + 1 ) + 2 e 4 ! ( e 2 + 1 ) + 2 e 6 ! ( e 2 + 1 ) + 2 e 8 ! ( e 2 + 1 ) + ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {2e}{e^{2}+1}}{\frac {1}{(2n)!}}={\frac {2e}{e^{2}+1}}+{\frac {2e}{2!(e^{2}+1)}}+{\frac {2e}{4!(e^{2}+1)}}+{\frac {2e}{6!(e^{2}+1)}}+{\frac {2e}{8!(e^{2}+1)}}+\cdots =1}
∑ n = 0 ∞ ( − 1 ) n ( 2 n ) ! cos 1 = 1 cos 1 − 1 2 ! cos 1 + 1 4 ! cos 1 − 1 6 ! cos 1 + 1 8 ! cos 1 − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!\cos 1}}={\frac {1}{\cos 1}}-{\frac {1}{2!\cos 1}}+{\frac {1}{4!\cos 1}}-{\frac {1}{6!\cos 1}}+{\frac {1}{8!\cos 1}}-\cdots =1}
∑ n = 0 ∞ 1 n ! e = 1 e + 1 e + 1 2 e + 1 6 e + 1 24 e + 1 120 e + ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n!e}}={\frac {1}{e}}+{\frac {1}{e}}+{\frac {1}{2e}}+{\frac {1}{6e}}+{\frac {1}{24e}}+{\frac {1}{120e}}+\cdots =1}
∑ n = 0 ∞ ( − 1 ) n n ! e = e − e + e 2 − e 6 + e 24 − e 120 + ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}e=e-e+{\frac {e}{2}}-{\frac {e}{6}}+{\frac {e}{24}}-{\frac {e}{120}}+\cdots =1}
∑ n = 0 ∞ − ( − 1 ) n π 2 n + 1 ( 2 n + 1 ) ! = − 1 + π 3 3 ! − π 5 5 ! + π 7 7 ! − π 9 9 ! + ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }-{\frac {(-1)^{n}\pi ^{2n+1}}{(2n+1)!}}=-1+{\frac {\pi ^{3}}{3!}}-{\frac {\pi ^{5}}{5!}}+{\frac {\pi ^{7}}{7!}}-{\frac {\pi ^{9}}{9!}}+\cdots =1}
∑ n = 0 ∞ ( − 1 ) n π 2 n + 1 ( 2 n + 1 ) ! 2 4 n + 1.5 = 1 − π 3 3 ! 2 5.5 + π 5 5 ! 2 9.5 − π 7 7 ! 2 13.5 + π 9 9 ! 2 17.5 − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}\pi ^{2n+1}}{(2n+1)!2^{4n+1.5}}}=1-{\frac {\pi ^{3}}{3!2^{5.5}}}+{\frac {\pi ^{5}}{5!2^{9.5}}}-{\frac {\pi ^{7}}{7!2^{13.5}}}+{\frac {\pi ^{9}}{9!2^{17.5}}}-\cdots =1}
∑ n = 0 ∞ ( − 1 ) n π 2 n ( 2 n ) ! 2 4 n − 1 2 = 1 − π 2 2 ! 2 3.5 + π 4 4 ! 2 7.5 − π 6 6 ! 2 11.5 + π 8 8 ! 2 15.5 − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}\pi ^{2n}}{(2n)!2^{4n-{\frac {1}{2}}}}}=1-{\frac {\pi ^{2}}{2!2^{3.5}}}+{\frac {\pi ^{4}}{4!2^{7.5}}}-{\frac {\pi ^{6}}{6!2^{11.5}}}+{\frac {\pi ^{8}}{8!2^{15.5}}}-\cdots =1}
∑ n = 0 ∞ − ( i π ) n n ! = − 1 − i π + π 2 2 + i π 3 6 − π 4 24 − i π 5 120 + ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }-{\frac {(i\pi )^{n}}{n!}}=-1-i\pi +{\frac {\pi ^{2}}{2}}+{\frac {i\pi ^{3}}{6}}-{\frac {\pi ^{4}}{24}}-{\frac {i\pi ^{5}}{120}}+\cdots =1}
∑ n = 0 ∞ 2 i n 1 + i = 2 1 + i + 2 i 1 + i − 2 1 + i − 2 i 1 + i + 2 1 + i + ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {2i^{n}}{1+i}}={\frac {2}{1+i}}+{\frac {2i}{1+i}}-{\frac {2}{1+i}}-{\frac {2i}{1+i}}+{\frac {2}{1+i}}+\cdots =1}
∑ n = 0 ∞ 2 i − n 1 − i = 2 1 − i − 2 i 1 − i − 2 1 − i + 2 i 1 − i + 2 1 − i − ⋯ = 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {2i^{-n}}{1-i}}={\frac {2}{1-i}}-{\frac {2i}{1-i}}-{\frac {2}{1-i}}+{\frac {2i}{1-i}}+{\frac {2}{1-i}}-\cdots =1}
