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Question Number 158522 by amin96 last updated on 05/Nov/21

$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{(\zeta \left(n\right)-n)}{2^n}=?$$

Answered by qaz last updated on 06/Nov/21

$$\underset{\mathrm{n}=2}{\sum ^{\mathrm{\infty }}}\frac{\zeta \left(\mathrm{n}\right)-\mathrm{n}}{2^{\mathrm{n}}}$$$$=\underset{\mathrm{n}=2}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\left(2\mathrm{k}\right)}^{\mathrm{n}}}-\frac{3}{2}$$$$=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{\frac{1}{{\left(2\mathrm{k}\right)}^2}}{1-\frac{1}{2\mathrm{k}}}-\frac{3}{2}$$$$=\frac{1}{2}\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{\mathrm{k}-\frac{1}{2}}-\frac{1}{\mathrm{k}})-\frac{3}{2}$$$$=-\frac{1}{2}{\mathrm{H}}_{-1/2}-\frac{3}{2}$$$$=\ln 2-\frac{3}{2}$$

Commented by amin96 last updated on 06/Nov/21

$$thankssir.verygoodsolution$$

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