Степенен ред

от **Гост** » 16 Апр 2019, 11:33

Определете областта на сходимост на степенния ред [tex]\sum_{n=1}^{\infty}\frac{x^{4n}}{(n+1)(3^{n}+2}[/tex].

$x\in(-\sqrt[4]{3};\sqrt[4]{3})$

Интересно е да се отбележи, че за $x=\pm\sqrt[4]{3}$ по Раабе-Дюамел се оказва разходящ.

от **Гост** » 17 Апр 2019, 06:39

Добромир Глухаров написа:<math><semantics><mrow><mstyle scriptlevel="0" displaystyle="true"><mi>x</mi><mo>∈</mo><mo>(</mo><mo>−</mo><mroot><mn>3</mn><mn>4</mn></mroot><mo separator="true">;</mo><mroot><mn>3</mn><mn>4</mn></mroot><mo>)</mo></mstyle></mrow><annotation encoding="application/x-tex">\displaystyle x\in(-\sqrt[4]{3};\sqrt[4]{3})</annotation></semantics></math>x∈(−43<svg width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
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Интересно е да се отбележи, че за <math><semantics><mrow><mstyle scriptlevel="0" displaystyle="true"><mi>x</mi><mo>=</mo><mo>±</mo><mroot><mn>3</mn><mn>4</mn></mroot></mstyle></mrow><annotation encoding="application/x-tex">\displaystyle x=\pm\sqrt[4]{3}</annotation></semantics></math>x=±43<svg width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
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s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z"></path></svg> по Раабе-Дюамел се оказва разходящ.

Може ли да кажеш как става, понеже ще имам задача от този тип на контролно?

$u_n=\frac{x^{4n}}{(n+1)(3^n+2)}$

Сходимост по Коши:

$\sqrt[n]{u_n}=\frac{x^4}{\sqrt[n]{n+1}\sqrt[n]{3^n+2}}\to\frac{x^4}{3}<1\Rightarrow|x|<\sqrt[4]{3}$

За $x=\pm\sqrt[4]{3}\to u_n=\frac{3^n}{(n+1)(3^n+2)}$

По Раабе-Дюамел:

$n\left(1-\frac{u_{n+1}}{u_n}\right)=n\left(1-\frac{3(n+1)(3^n+2)}{(n+2)(3^{n+1}+2)}\right)=$

$=n\left(1-\frac{n.3^{n+1}+3^{n+1}+6n+6}{n.3^{n+1}+2.3^{n+1}+2n+4}\right)=$

$=n\cdot\frac{n.3^{n+1}+2.3^{n+1}+2n+4-n.3^{n+1}-3^{n+1}-6n-6}{n.3^{n+1}+2.3^{n+1}+2n+4}=$

$=\frac{n.3^{n+1}-4n^2-2n}{n.3^{n+1}+2.3^{n+1}+2n+4}<1$

$\Rightarrow$ за $x=\pm\sqrt[4]{3}$ е разходящ.

Степенен ред

Степенен ред

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Кой е на линия