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Question Number 145960 by puissant last updated on 10/Jul/21

Answered by mathmax by abdo last updated on 10/Jul/21

$${\mathrm{R}}_{\mathrm{n}}=\sum _{\mathrm{p}=\mathrm{n}+1}^{2\mathrm{n}}\sin \left(\frac{1}{\mathrm{p}}\right)\Rightarrow {\mathrm{R}}_{\mathrm{n}}{=}_{\mathrm{p}-\mathrm{n}=\mathrm{k}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\sin \left(\frac{1}{\mathrm{n}+\mathrm{k}}\right)$$$$\mathrm{we}\mathrm{have}\mathrm{sinx}=\mathrm{x}-\frac{{\mathrm{x}}^3}{6}+...\Rightarrow \mathrm{x}-\frac{{\mathrm{x}}^3}{6}⩽\mathrm{sinx}⩽\mathrm{x}\Rightarrow$$$$\frac{1}{\mathrm{n}+\mathrm{k}}-\frac{1}{6{(\mathrm{n}+\mathrm{k})}^3}⩽\sin \left(\frac{1}{\mathrm{n}+\mathrm{k}}\right)⩽\frac{1}{\mathrm{n}+\mathrm{k}}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{n}+\mathrm{k}}-\frac{1}{6}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{{(\mathrm{n}+\mathrm{k})}^3}⩽\sum _{\mathrm{k}=1}^{\mathrm{n}}\sin \left(\frac{1}{\mathrm{n}+\mathrm{k}}\right)⩽\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{n}+\mathrm{k}}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{n}+\mathrm{k}}=\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{1+\frac{\mathrm{k}}{\mathrm{n}}}\to {\int }_0^1\frac{\mathrm{dx}}{1+\mathrm{x}}=\log 2$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{{(\mathrm{n}+\mathrm{k})}^3}\mathrm{we}\mathrm{k}⩾1\Rightarrow \mathrm{n}+\mathrm{k}⩾\mathrm{n}+1\Rightarrow \frac{1}{{(\mathrm{n}+\mathrm{k})}^3}⩽\frac{1}{{(\mathrm{n}+1)}^3}$$$$\Rightarrow \sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{{(\mathrm{n}+\mathrm{k})}^3}⩽\frac{\mathrm{n}}{{(\mathrm{n}+1)}^3}\to 0(\mathrm{n}\to \mathrm{\infty })\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{R}}_{\mathrm{n}}=\log 2$$

Commented by mathmax by abdo last updated on 10/Jul/21

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}.$$

Commented by puissant last updated on 10/Jul/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by puissant last updated on 10/Jul/21

$$\mathrm{okey}\mathrm{thanks}$$

Answered by mathmax by abdo last updated on 10/Jul/21

$${\mathrm{A}}_{\mathrm{n}}=\sum _{\mathrm{p}=\mathrm{n}+1}^{2\mathrm{n}}\frac{1}{{\mathrm{p}}^{\alpha }}\Rightarrow {\mathrm{A}}_{\mathrm{n}}{=}_{\mathrm{p}-\mathrm{n}=\mathrm{k}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{{(\mathrm{n}+\mathrm{k})}^{\alpha }}$$$$\mathrm{n}+\mathrm{k}>\mathrm{n}\Rightarrow \frac{1}{{(\mathrm{n}+\mathrm{k})}^{\alpha }}<\frac{1}{{\mathrm{n}}^{\alpha }}\Rightarrow {\mathrm{A}}_{\mathrm{n}}<\frac{\mathrm{n}}{{\mathrm{n}}^{\alpha }}=\frac{1}{{\mathrm{n}}^{\alpha -1}}$$$$\mathrm{but}\alpha -1>0\Rightarrow {\lim }_{\mathrm{n}\to \mathrm{\infty }}\frac{1}{{\mathrm{n}}^{\alpha -1}}=0\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}=0$$

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