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Question Number 117787 by mathdave last updated on 13/Oct/20

Answered by Dwaipayan Shikari last updated on 13/Oct/20

$$\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^1{x}^n\frac{1-x^n}{1-x}dx$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^1{x}^n(1+x+x^2+...x^{n-1})dx$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}$$$$=\frac{1}{n}\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}\frac{1}{1+\frac{k}{n}}$$$$={\int }_0^1\frac{1}{1+x}dx={\left[log(1+x)\right]}_0^1=log\left(2\right)$$

Answered by mathmax by abdo last updated on 13/Oct/20

$${\mathrm{A}}_{\mathrm{n}}={\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}-{\mathrm{x}}^{2\mathrm{n}}}{1-\mathrm{x}}\mathrm{dx}\Rightarrow {\mathrm{A}}_{\mathrm{n}}={\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}(1-{\mathrm{x}}^{\mathrm{n}})}{1-\mathrm{x}}\mathrm{dx}$$$$={\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}(1-\mathrm{x})(1+\mathrm{x}+{\mathrm{x}}^2+...+{\mathrm{x}}^{\mathrm{n}-1})}{1-\mathrm{x}}\mathrm{dx}$$$$={\int }_0^1({\mathrm{x}}^{\mathrm{n}}+{\mathrm{x}}^{\mathrm{n}+1}+{\mathrm{x}}^{\mathrm{n}+2}+...+{\mathrm{x}}^{2\mathrm{n}-1})\mathrm{dx}$$$$={[\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\frac{{\mathrm{x}}^{\mathrm{n}+2}}{\mathrm{n}+2}+....+\frac{{\mathrm{x}}^{2\mathrm{n}}}{2\mathrm{n}}]}_0^1=\frac{1}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+2}+....+\frac{1}{\mathrm{n}+\mathrm{n}}$$$$=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{n}+\mathrm{k}}\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{1}{\mathrm{n}}{\sum _{\mathrm{k}=1}}^{\mathrm{n}}\frac{1}{1+\frac{\mathrm{k}}{\mathrm{n}}}$$$$={\int }_0^1\frac{\mathrm{dx}}{1+\mathrm{x}}={\left[\ln (1+\mathrm{x})\right]}_0^1=\ln \left(2\right)$$

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