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Question Number 201860 by MrGHK last updated on 14/Dec/23

$$\underset{\mathbf{n}=1}{\mathbf{\sum }^{\mathrm{\infty }}}\frac{{(-1)}^{\mathbf{n}}{\mathbf{H}}_{\mathbf{n}}}{\mathbf{n}+1}=??$$

Answered by mnjuly1970 last updated on 14/Dec/23

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{H}_n{x}^n=-\frac{ln(1-x)}{1-x}$$$$where,H_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2}$$$${\int }_0^x\underset{n=1}{\sum ^{\mathrm{\infty }}}{H}_n{t}^{n}dt=\frac{1}{2}{ln}^2(1-x)+C$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n{x}^{n+1}}{n+1}=\frac{1}{2}{ln}^2(1-x)+C$$$$x=0\Rightarrow C=0$$$$\underset{n=}{\sum ^{\mathrm{\infty }}}\frac{H_n{x}^{n+1}}{n+1}=\frac{1}{2}{ln}^2(1-x)$$$$x=-1\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{H}_n}{n+1}=-\frac{1}{2}{ln}^2\left(2\right)$$$$$$

Commented by MrGHK last updated on 14/Dec/23

$$\mathbf{so}\mathbf{we}\mathbf{can}\mathbf{say}\underset{\mathbf{n}=1}{\mathbf{\sum }^{\mathrm{\infty }}}\frac{{\mathbf{H}}_{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}+1}}{\mathbf{n}+1}=\frac{1}{2}{\mathbf{ln}}^2(1-\mathbf{x})$$

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