Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 203329 by Calculusboy last updated on 16/Jan/24

$$\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{t}\mathit{h}\mathit{e}\mathit{r}\mathit{a}\mathit{n}\mathit{g}\mathit{e}\mathit{s}\mathit{o}\mathit{f}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{e}\mathit{o}\mathit{f}\mathit{x}\mathit{f}\mathit{o}\mathit{r}\mathit{w}\mathit{h}\mathit{i}\mathit{c}\mathit{h}$$$$\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}\mathit{c}\mathit{o}\mathit{n}\mathit{v}\mathit{e}\mathit{r}\mathit{g}\mathit{e}\mathit{n}\mathit{t}\mathit{o}\mathit{r}\mathit{d}\mathit{i}\mathit{v}\mathit{e}\mathit{r}\mathit{g}\mathit{e}\mathit{n}\mathit{t}$$$$\underset{\mathit{n}=1}{\mathbf{\sum }^{\mathrm{\infty }}}\frac{(\mathit{n}+1)}{{\mathit{n}}^3}{\mathit{x}}^{\mathit{n}}$$

Answered by Mathspace last updated on 17/Jan/24

$$u_n\left(x\right)=\frac{n+1}{n^3}{x}^n$$$$\mid \frac{u_{n+1}\left(x\right)}{u_n\left(x\right)}\mid =\mid \frac{\frac{n+2}{{(n+1)}^3}{x}^{n+1}}{\frac{n+1}{n^3}{x}^n}\mid$$$$=\frac{n+2}{{(n+1)}^3}\times \frac{n^3}{n+1}\mid x\mid$$$$={\left(\frac{n}{n+1}\right)}^3.\frac{n+2}{n+1}\mid x\mid \to \mid x\mid (n\to +\mathrm{\infty })$$$$si\mid x\mid <1theseriecvandr=1$$$$si\mid x\mid >1theseriediverges$$$$ifx=1u_n\left(x\right)=\frac{n+1}{n^3}\sim \frac{1}{n^2}$$$$\mathrm{\Sigma }\frac{1}{n^2}cv\Rightarrow \mathrm{\Sigma }{u}_n\left(x\right)cv$$$$if=x=-1u_n\left(x\right)=\frac{(n+1)}{n^3}{(-1)}^n$$$$\mathrm{\Sigma }{u}_n\left(x\right)isaalternatingserie$$$$\Rightarrow \mathrm{\Sigma }{u}_n\left(x\right)cv.$$$$$$

Commented by Calculusboy last updated on 17/Jan/24

$$\mathit{n}\mathit{i}\mathit{c}\mathit{e}\mathit{s}\mathit{i}\mathit{r}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com