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Question Number 33354 by caravan msup abdo. last updated on 15/Apr/18

find the radius of Σ_(n≥1)  ((ln(n))/(√(n^3  +n+1))) z^n

$${find}\:{the}\:{radius}\:{of}\:\sum_{{n}\geqslant\mathrm{1}} \:\frac{{ln}\left({n}\right)}{\sqrt{{n}^{\mathrm{3}} \:+{n}+\mathrm{1}}}\:{z}^{{n}} \\ $$

Commented by math khazana by abdo last updated on 20/Apr/18

this serie is positifs terms let put u_(n ) =((ln(n))/(√(n^3  +n+1)))  u_n = ((ln(n))/((√n^3 )((√(1 +(1/n^2 ) +(1/n^3 )))))) ∼ ((ln(n))/(n^(3/2) ( 1+(1/2)((1/n^2 ) +(1/n^3 )))))  ∼    ((ln(n))/n^(3/2) ) =v_n     we have (v_(n+1) /v_n ) =  (((ln(n+1))/((n+1)^(3/2) ))/((ln(n))/n^(3/2) ))  = ((ln(n+1))/(ln(n))) ((n/(n+1)))^(3/2) →1(n→+∞) so the radius of  convergence is R =1 .

$${this}\:{serie}\:{is}\:{positifs}\:{terms}\:{let}\:{put}\:{u}_{{n}\:} =\frac{{ln}\left({n}\right)}{\sqrt{{n}^{\mathrm{3}} \:+{n}+\mathrm{1}}} \\ $$$${u}_{{n}} =\:\frac{{ln}\left({n}\right)}{\sqrt{{n}^{\mathrm{3}} }\left(\sqrt{\mathrm{1}\:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{n}^{\mathrm{3}} }}\right)}\:\sim\:\frac{{ln}\left({n}\right)}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\right)\right)} \\ $$$$\sim\:\:\:\:\frac{{ln}\left({n}\right)}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:={v}_{{n}} \:\:\:\:{we}\:{have}\:\frac{{v}_{{n}+\mathrm{1}} }{{v}_{{n}} }\:=\:\:\frac{\frac{{ln}\left({n}+\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }}{\frac{{ln}\left({n}\right)}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }} \\ $$$$=\:\frac{{ln}\left({n}+\mathrm{1}\right)}{{ln}\left({n}\right)}\:\left(\frac{{n}}{{n}+\mathrm{1}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \rightarrow\mathrm{1}\left({n}\rightarrow+\infty\right)\:{so}\:{the}\:{radius}\:{of} \\ $$$${convergence}\:{is}\:{R}\:=\mathrm{1}\:. \\ $$

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