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Question Number 122626 by rs4089 last updated on 18/Nov/20

Answered by Dwaipayan Shikari last updated on 18/Nov/20

$$S=1-1+1-1+1-1+...=\frac{1}{1+1}=\frac{1}{2}$$$$Butactuallysumdoesnotexist$$$$\zeta (1-s)=2^{1-s}{\pi }^{-s}cos\left(\frac{\pi s}{2}\right)\mathrm{\Gamma }\left(s\right)\zeta \left(s\right)$$$$\zeta (-1)=-\frac{1}{2{\pi }^2}\mathrm{\Gamma }\left(2\right)\zeta \left(2\right)=-\frac{1}{12}\left(strangeenough\right)$$$$\zeta (-1)=1+2+3+4+5+6+7+....$$

Answered by mathmax by abdo last updated on 18/Nov/20

$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}=1-1+1-1+....=\frac{1}{2}$$

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