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Question Number 31672 by gunawan last updated on 12/Mar/18

$$\mathrm{for}\mathrm{what}\mathrm{value}p\mathrm{is}\mathrm{a}\mathrm{series}$$$$\text{Missing left or extra right}$$

Commented by abdo imad last updated on 12/Mar/18

$$sinx=x-\frac{x^3}{6}+o\left(x^5\right)\Rightarrow sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{6n^3}+o\left(\frac{1}{n^5}\right)(n\to \mathrm{\infty })$$$$\Rightarrow -sin\left(\frac{1}{n}\right)=-\frac{1}{n}+\frac{1}{6n^3}+o\left(\frac{1}{n^5}\right)\Rightarrow \frac{1}{n}-sin\left(\frac{1}{n}\right)=\frac{1}{6n^3}+o\left(\frac{1}{n^5}\right)\Rightarrow$$$${(\frac{1}{n}-sin\left(\frac{1}{n}\right))}^p\sim \frac{1}{6n^{3p}}andtheserie\sum _{p⩾1}\frac{1}{6n^{3p}}converges$$$$if3p>1\iff p>\frac{1}{3}fpisintegrwemusthavep⩾1.$$

Commented by gunawan last updated on 12/Mar/18

$$\mathrm{thank}\mathrm{you}\mathrm{Sir}$$

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