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Question Number 102544 by 175mohamed last updated on 09/Jul/20

Answered by Rio Michael last updated on 10/Jul/20

$$\mathrm{for}p>1,\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^p}\mathrm{is}\mathrm{convergent}\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\sqrt[3]{n^2}}\mathrm{diverges}$$

Answered by mathmax by abdo last updated on 10/Jul/20

$$\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^{\frac{2}{3}}}\mathrm{the}\mathrm{function}\phi \left(\mathrm{t}\right)=\frac{1}{{\mathrm{t}}^{\frac{2}{3}}}\mathrm{are}>0\mathrm{decreazing}\mathrm{so}\mathrm{S}\mathrm{have}\mathrm{same}$$$$\mathrm{nature}\mathrm{with}{\int }_1^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^{\frac{2}{3}}}\mathrm{this}\mathrm{integral}\mathrm{is}\mathrm{divergent}\Rightarrow \mathrm{S}\mathrm{is}\mathrm{divergent}$$$$\mathrm{any}\mathrm{way}\mathrm{\Sigma }\frac{1}{{\mathrm{n}}^{\alpha }}\mathrm{converge}\mathrm{for}\mathrm{n}>1\mathrm{and}\mathrm{diverges}\mathrm{for}\alpha ⩽1$$

Commented by 175mohamed last updated on 13/Jul/20

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