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Question Number 66483 by ~ À ® @ 237 ~ last updated on 16/Aug/19

$$calculate\underset{k=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k}{k}\zeta \left(k\right)if\zeta \left(s\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^s}$$

Commented by mathmax by abdo last updated on 16/Aug/19

$$letS=\sum _{k=2}^{\mathrm{\infty }}\frac{{(-1)}^k}{k}\xi \left(k\right)\Rightarrow S=\sum _{k=2}^{\mathrm{\infty }}\frac{{(-1)}^k}{k}\left(\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^k}\right)$$$$=\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{k=2}^{\mathrm{\infty }}\frac{{(-1)}^k}{k{n}^k}\right)=\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{k=2}^{\mathrm{\infty }}\frac{1}{k}{(-\frac{1}{n})}^k\right)$$$$letfindf\left(x\right)=\sum _{k=2}^{\mathrm{\infty }}\frac{x^k}{k}with\mid x\mid <1\Rightarrow f^{{}^{\prime }}\left(x\right)=\sum _{k=2}^{\mathrm{\infty }}{x}^{k-1}=\sum _{k=1}^{\mathrm{\infty }}{x}^k$$$$=x\sum _{k=1}^{\mathrm{\infty }}{x}^{k-1}[=x\sum _{k=0}^{\mathrm{\infty }}{x}^k=\frac{x}{1-x}\Rightarrow f\left(x\right)=\int \frac{x}{1-x}dx+c$$$$=-\int \frac{x}{x-1}dx+c=-\int \frac{x-1+1}{x-1}dx+c=-x+ln\mid x-1\mid +c$$$$f\left(0\right)=0=c\Rightarrow f\left(x\right)=-x+ln\mid x-1\mid \Rightarrow$$$$\sum _{k=2}^{\mathrm{\infty }}\frac{1}{k}{(-\frac{1}{n})}^k=\frac{1}{n}+ln\mid \frac{1}{n}+1\mid =\frac{1}{n}+ln\left(\frac{n+1}{n}\right)$$$$=\frac{1}{n}+ln\left(n\right)-ln(n+1)\Rightarrow$$$$S=\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{n}+ln\left(n\right)-ln(n+1))={lim}_{n\to +\mathrm{\infty }}{S}_{n}with$$$$S_n=\sum _{k=1}^n(\frac{1}{k}+ln\left(k\right)-ln(k+1))=\sum _{k=1}^n\frac{1}{k}+\sum _{k=1}^n(ln\left(k\right)-ln(k+1))$$$$=H_n-ln(n+1)=H_n-ln\left(n\right)-ln(1+\frac{1}{n})\to \gamma (n\to +\mathrm{\infty })\Rightarrow$$$$\sum _{k=2}^{\mathrm{\infty }}\frac{{(-1)}^k}{k}\xi \left(k\right)=\gamma \left(constantofEuler\right)$$

Commented by mathmax by abdo last updated on 17/Aug/19

$$f\left(x\right)=-x-ln\mid x-1\mid \Rightarrow \sum _{k=2}^{\mathrm{\infty }}\frac{1}{k}{(-\frac{1}{n})}^k=\frac{1}{n}-ln(\frac{1}{n}+1)$$$$=\frac{1}{n}-ln\left(\frac{n+1}{n}\right)=\frac{1}{n}+ln\left(n\right)-ln(n+1)...\left(erroroftypo\right)$$

Commented by ~ À ® @ 237 ~ last updated on 17/Aug/19

$$thankyousir$$

Commented by mathmax by abdo last updated on 17/Aug/19

$$youarewelcome.$$

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