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Question Number 131590 by pticantor last updated on 06/Feb/21

$$\mathit{s}\mathit{h}\mathit{o}\mathit{w}\mathit{t}\mathit{h}\mathit{a}\mathit{t}$$$${\mathit{U}}_{\mathit{n}}={\int }_0^1\frac{\mathit{d}\mathit{x}}{1+\mathit{x}+{\mathit{x}}^2+....+{\mathit{x}}^{\mathit{n}}}$$$$\mathit{c}\mathit{o}\mathit{n}\mathit{v}\mathit{e}\mathit{r}\mathit{g}\mathit{e}\mathit{s}!!$$

Answered by mathmax by abdo last updated on 06/Feb/21

$${\mathrm{U}}_{\mathrm{n}}={\int }_0^1\frac{(1-\mathrm{x})}{1-{\mathrm{x}}^{\mathrm{n}+1}}\mathrm{dx}\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{U}}_{\mathrm{n}}={\int }_0^1{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{1-\mathrm{x}}{1-{\mathrm{x}}^{\mathrm{n}+1}}\mathrm{dx}$$$$={\int }_0^1(1-\mathrm{x})\mathrm{dx}={[\mathrm{x}-\frac{{\mathrm{x}}^2}{2}]}_0^1=1-\frac{1}{2}=\frac{1}{2}\mathrm{so}{\mathrm{U}}_{\mathrm{n}}\mathrm{cv}\mathrm{and}{\mathrm{U}}_{\mathrm{n}}\to \frac{1}{2}$$

Commented by pticantor last updated on 08/Feb/21

$$\mathit{i}\mathit{t}\mathit{h}\mathit{i}\mathit{n}\mathit{k}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathit{s}\mathit{h}\mathit{o}\mathit{u}\mathit{l}\mathit{d}\mathit{b}\mathit{e}\mathit{t}\mathit{r}\mathit{u}\mathit{e}\mathit{i}\mathit{f}$$$$\mid \mathit{x}\mid <1$$$$\mathit{i}\mathit{h}\mathit{a}\mathit{v}\mathit{e}\mathit{a}\mathit{q}\mathit{u}\mathit{e}\mathit{s}\mathit{t}\mathit{i}\mathit{o}\mathit{n}!\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}\mathit{i}\mathit{n}\mathit{w}\mathit{i}\mathit{c}\mathit{h}\mathit{c}\mathit{a}\mathit{s}\mathit{e}\mathit{s}$$$$\mathit{w}\mathit{e}\mathit{h}\mathit{a}\mathit{v}\mathit{e}$$$$\text{Double subscripts: use braces to clarify}$$

Answered by Dwaipayan Shikari last updated on 06/Feb/21

$${\int }_0^1\frac{1-x}{1-x^{n+1}}dx{x}^{n+1}=u$$$$\Rightarrow \frac{1}{n+1}{\int }_0^1\frac{x^{-n}-x^{1-n}}{1-u}du=\frac{1}{n+1}{\int }_0^1\frac{u^{-\frac{n}{n+1}}-u^{\frac{1-n}{n+1}}}{1-u}du$$$$=\frac{1}{n+1}(\psi (1+\frac{1-n}{n+1})-\psi (1-\frac{n}{n+1}))$$$$=\frac{1}{n+1}(\psi \left(\frac{2}{n+1}\right)-\psi \left(\frac{1}{n+1}\right))$$

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