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Question-102544

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

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Question-102545

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Question Number 102545 by 175mohamed last updated on 09/Jul/20 Answered by mathmax by abdo last updated on 13/Jul/20 $$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{n}^{\mathrm{1},\mathrm{2}} −\mathrm{1}}{\mathrm{n}^{\mathrm{2},\mathrm{3}} +\mathrm{1}}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{n}^{\frac{\mathrm{6}}{\mathrm{5}}}…

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