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Question Number 77412 by BK last updated on 06/Jan/20

Commented by Tony Lin last updated on 06/Jan/20

$$\zeta \left(s\right)=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\cdot \cdot \cdot +\frac{1}{n^s}$$$$\zeta \left(2\right)=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdot \cdot \cdot +\frac{1}{n^2}=\frac{{\pi }^2}{6}$$$$\frac{1}{2^2}+\frac{1}{3^2}+\cdot \cdot \cdot +\frac{1}{n^2}=\frac{{\pi }^2}{6}-1$$

Commented by mr W last updated on 06/Jan/20

$$\zeta \left(s\right)=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\cdot \cdot \cdot +\frac{1}{n^s}+......$$$$\zeta \left(s\right)\ne \frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\cdot \cdot \cdot +\frac{1}{n^s}$$$$\frac{1}{2^2}+\frac{1}{3^2}+\cdot \cdot \cdot +\frac{1}{n^2}+.....=\frac{{\pi }^2}{6}-1$$$$\frac{1}{2^2}+\frac{1}{3^2}+\cdot \cdot \cdot +\frac{1}{n^2}\ne \frac{{\pi }^2}{6}-1$$

Commented by BK last updated on 06/Jan/20

$$\mathrm{thanks}$$

Commented by Tony Lin last updated on 06/Jan/20

$$sorryIgotit$$

Commented by BK last updated on 06/Jan/20

$$\frac{{\pi }^2}{6}-\mathrm{prove}\mathrm{that}\mathrm{sir}\mathrm{plz}$$

Commented by mathmax by abdo last updated on 06/Jan/20

$$notcorrect\xi \left(s\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^s}\left(itsainfinitesum\right)$$$$and\frac{1}{1^s}+\frac{1}{2^s}+....+\frac{1}{n^s}\ne \xi \left(s\right)$$

Commented by mathmax by abdo last updated on 07/Jan/20

$$S=\sum _{k=1}^n\frac{1}{k^2}-1=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}-1-\sum _{k=n+1}^{\mathrm{\infty }}\frac{1}{k^2}$$$$=\frac{{\pi }^2}{6}-1-\sum _{k=n+1}^{\mathrm{\infty }}\frac{1}{k^2}$$

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