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Question Number 142290 by mnjuly1970 last updated on 29/May/21

$$evaluate:$$$$\mathrm{\Theta }:=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\zeta (n+1)-1}{n+1}=?$$$$$$

Answered by Dwaipayan Shikari last updated on 29/May/21

$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{\zeta \left(n\right)-1}{n}=\underset{n=2}{\sum ^{\mathrm{\infty }}}\underset{k=2}{\sum ^{\mathrm{\infty }}}\frac{1}{{nk}^n}$$$$=\underset{k=2}{\sum ^{\mathrm{\infty }}}(-log(1-\frac{1}{k})-\frac{1}{k})$$$$=\underset{n\to \mathrm{\infty }}{\lim }-log(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1})-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}+1$$$$=log(\frac{2}{1}.\frac{3}{2}.\frac{4}{3}....\frac{n}{n-1}.\frac{n+1}{n})-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}+1$$$$=\underset{n\to \mathrm{\infty }}{\lim }log(n+1)-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}+1=1-\gamma$$$$$$

Commented by mnjuly1970 last updated on 29/May/21

$$thanksalot...$$

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