Question-102545

Question Number 102545 by 175mohamed last updated on 09/Jul/20

Answered by mathmax by abdo last updated on 13/Jul/20

Σ_(n=1) ^∞ ((n^(1,2) −1)/(n^(2,3) +1)) =Σ_(n=1) ^∞ ((n^(6/5) −1)/(n^((23)/(10)) +1)) =Σ_(n=1) ^∞ U_n U_n ∼ (n^(6/5) /n^((23)/(10)) ) =(1/n^(((23)/(10))−(6/5)) ) =(1/n^((23−12)/(10)) ) =(1/n^((11)/(10)) ) and Σ(1/n^((11)/(10)) ) is convergent (((11)/(10))>1) ⇒Σ U_n converges

$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{n}^{\mathrm{1},\mathrm{2}} −\mathrm{1}}{\mathrm{n}^{\mathrm{2},\mathrm{3}} +\mathrm{1}}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{n}^{\frac{\mathrm{6}}{\mathrm{5}}} −\mathrm{1}}{\mathrm{n}^{\frac{\mathrm{23}}{\mathrm{10}}} \:+\mathrm{1}}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{U}_{\mathrm{n}} \\ $$$$\mathrm{U}_{\mathrm{n}} \sim\:\frac{\mathrm{n}^{\frac{\mathrm{6}}{\mathrm{5}}} }{\mathrm{n}^{\frac{\mathrm{23}}{\mathrm{10}}} }\:=\frac{\mathrm{1}}{\mathrm{n}^{\frac{\mathrm{23}}{\mathrm{10}}−\frac{\mathrm{6}}{\mathrm{5}}} }\:=\frac{\mathrm{1}}{\mathrm{n}^{\frac{\mathrm{23}−\mathrm{12}}{\mathrm{10}}} }\:=\frac{\mathrm{1}}{\mathrm{n}^{\frac{\mathrm{11}}{\mathrm{10}}} }\:\:\mathrm{and}\:\Sigma\frac{\mathrm{1}}{\mathrm{n}^{\frac{\mathrm{11}}{\mathrm{10}}} }\:\mathrm{is}\:\mathrm{convergent}\:\left(\frac{\mathrm{11}}{\mathrm{10}}>\mathrm{1}\right)\:\Rightarrow\Sigma\:\mathrm{U}_{\mathrm{n}} \mathrm{converges} \\ $$

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Question-102545

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