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Question Number 145665 by phally last updated on 07/Jul/21

Answered by puissant last updated on 07/Jul/21

$${\mathrm{k}}^{\prime }=2\mathrm{k}-1\Rightarrow 1⩽{\mathrm{k}}^{\prime }⩽2\mathrm{n}-1$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\underset{{\mathrm{k}}^{\prime }=1}{\sum ^{2\mathrm{n}-1}}\frac{2}{2\mathrm{n}+{\mathrm{k}}^{\prime }}$$$$={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{2}{2\mathrm{n}+\mathrm{k}}+{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\underset{\mathrm{k}=\mathrm{n}+1}{\sum ^{2\mathrm{n}-1}}\frac{2}{2\mathrm{n}+\mathrm{k}}$$$$={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{1}{\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{2}{2+\left(\frac{\mathrm{k}}{\mathrm{n}}\right)}+{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{1}{\mathrm{n}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}\frac{2}{2+\left(\frac{\mathrm{k}}{\mathrm{n}}\right)}$$$$=2{\int }_0^1\frac{1}{2+\mathrm{x}}\mathrm{dx}+2{\int }_0^1\frac{1}{2+\mathrm{x}}\mathrm{dx}$$$$=4{\int }_0^1\frac{1}{2+\mathrm{x}}\mathrm{dx}=4{\left[\ln (2+\mathrm{x})\right]}_0^1=4(\ln 3-\ln 2)$$$$\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{2}{2\mathrm{n}+2\mathrm{k}-1}=4\ln \left(\frac{3}{2}\right)..$$

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