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Question Number 146767 by qaz last updated on 15/Jul/21

$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}(1+\frac{1}{3}+...+\frac{1}{2\mathrm{n}+1})=?$$

Answered by mnjuly1970 last updated on 16/Jul/21

$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n+1}\{{\mathrm{H}}_{2n+1}-\frac{1}{2}{\mathrm{H}}_n\}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{\mathrm{H}}_{2n+1}}{n+1}-\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{\mathrm{H}}_n}{n+1}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{\mathrm{H}}_{2n+1}}{n+1}-\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{\mathrm{H}}_n}{n+1}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{\mathrm{H}}_{2n+1}}{n+1}+\frac{1}{4}{\log }^2\left(2\right)$$$$=\{\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{\mathrm{H}}_{2n+1}}{n+1}=\mathrm{S}\}+\frac{1}{4}{\log }^2\left(2\right)\left(★\right)$$$$$$$$\mathrm{S}:=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{{\mathrm{H}}_{2n+1}}{n+1}=?$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{H}}_n}{n+1}{x}^n=\frac{1}{2x}{\log }^2(1-x)$$$$x=i\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{H}}_n}{n+1}{i}^n=\frac{1}{2i}{\log }^2\left(\sqrt{2}{\mathrm{e}}^{-\frac{i\pi }{4}}\right)...\left(1\right)$$$$x=-i\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{H}}_n}{n+1}{(-i)}^n=\frac{1}{-2i}{\log }^2\left(\sqrt{2}{\mathrm{e}}^{\frac{i\pi }{4}}\right)+{\mathrm{Li}}_2(-\mathrm{i})...\left(2\right)$$$$\left(1\right)-\left(2\right)::\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n\left(2i\right){\mathrm{H}}_{2n+1}}{2n+2}=\frac{1}{2i}({\log }^2{\left(\sqrt{2}\right)}^{}-\frac{{\pi }^2}{16})$$$$=-\frac{1}{4}{\log }^2\left(2\right)+\frac{{\pi }^2}{16}\left(★★\right)$$$$$$$$\left(★★\right)\to \left(★\right).....\mathrm{A}nswer:=\frac{{\pi }^2}{16}◼$$$$icorrrctedmymistake$$$$\mathrm{S}irqaz....$$$$$$$$$$$$$$

Commented by qaz last updated on 16/Jul/21

$$\mathrm{Answer}\mathrm{is}\frac{{\pi }^2}{16}.\mathrm{Sir}$$

Commented by qaz last updated on 16/Jul/21

$$\mathrm{The}{\mathrm{other}}^{\prime }\mathrm{s}\mathrm{sol}....$$$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}(1+\frac{1}{3}+...+\frac{1}{2\mathrm{n}+1})$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1\frac{{(-1)}^{\mathrm{n}}(1-{\mathrm{x}}^{2\mathrm{n}+2})}{(\mathrm{n}+1)(1-{\mathrm{x}}^2)}\mathrm{dx}$$$$={\int }_0^1\frac{\ln 2-\ln (1+{\mathrm{x}}^2)}{1-{\mathrm{x}}^2}\mathrm{dx}$$$$={\int }_0^1{\int }_0^1\frac{1}{2-(1-{\mathrm{x}}^2)\mathrm{y}}\mathrm{dydx}$$$$={\int }_0^1\frac{{\tan }^{-1}\left(\sqrt{\frac{\mathrm{y}}{2-\mathrm{y}}}\right)}{\sqrt{2\mathrm{y}-{\mathrm{y}}^2}}\mathrm{dy}$$$$={\left({\tan }^{-1}\left(\sqrt{\frac{\mathrm{y}}{2-\mathrm{y}}}\right)\right)}^2{\mid }_0^1$$$$=\frac{{\pi }^2}{16}$$$$---------------------$$$$\mathrm{My}\mathrm{sol}...$$$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}(1+\frac{1}{3}+...+\frac{1}{2\mathrm{n}+1})$$$$={\int }_0^1\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\frac{{(-\mathrm{x})}^{\mathrm{n}}}{2\mathrm{k}+1}\mathrm{dx}$$$$={\int }_0^1\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{n}=\mathrm{k}}{\sum ^{\mathrm{\infty }}}\frac{{(-\mathrm{x})}^{\mathrm{n}}}{2\mathrm{k}+1}\mathrm{dx}$$$$={\int }_0^1\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-\mathrm{x})}^{\mathrm{n}+\mathrm{k}}}{2\mathrm{k}+1}\mathrm{dx}$$$$={\int }_0^1\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-\mathrm{x})}^{\mathrm{k}}}{2\mathrm{k}+1}\right)\left(\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-\mathrm{x})}^{\mathrm{n}}\right)\mathrm{dx}$$$$={\int }_0^1\frac{1}{1+\mathrm{x}}\left(\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{(-\mathrm{x})}^{\mathrm{k}}{\int }_0^1{\mathrm{y}}^{2\mathrm{k}}\mathrm{dy}\right)\mathrm{dx}$$$$={\int }_0^1\frac{1}{1+\mathrm{x}}{\int }_0^1\frac{1}{1+{\mathrm{xy}}^2}\mathrm{dydx}$$$$={\int }_0^1\frac{{\tan }^{-1}\sqrt{\mathrm{x}}}{(1+\mathrm{x})\sqrt{\mathrm{x}}}\mathrm{dx}$$$$=\frac{{\pi }^2}{16}$$

Commented by mnjuly1970 last updated on 16/Jul/21

$$thankyousomuch...$$

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