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Question Number 145450 by solihin last updated on 05/Jul/21

Answered by Olaf_Thorendsen last updated on 05/Jul/21

$$a)$$$$a_n=\frac{2^n}{3^{n+1}},n\in {\mathbb{N}}^{\ast }$$$$a_n=\frac{1}{3}{\left(\frac{2}{3}\right)}^n\underset{\mathrm{\infty }}{\to }0$$$$b)$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{n}^2}{3n^4+2n^3+5n-2}$$$$\mathrm{S}=\mathrm{\Sigma }{(-1)}^n{a}_n$$$$with\mid a_n\mid \underset{\mathrm{\infty }}{\to }0$$$$\mathrm{and}\mid \frac{a_{n+1}}{a_n}\mid <1$$$$\Rightarrow \mathrm{convergence}$$$$\mathrm{and}\mathrm{\Sigma }\mid a_n\mid =\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2}{3n^4+2n^3+5n-2}$$$$⩽\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2}{3n^4}=\frac{1}{3}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\frac{{\pi }^2}{18}$$$$\mathrm{S}\mathrm{is}\mathrm{absolutely}\mathrm{convergent}$$

Answered by mathmax by abdo last updated on 05/Jul/21

$${\mathrm{a}}_{\mathrm{n}}=\frac{2^{\mathrm{n}}}{3^{\mathrm{n}+1}}\Rightarrow {\mathrm{a}}_{\mathrm{n}}=\frac{1}{3}{\left(\frac{2}{3}\right)}^{\mathrm{n}}\mathrm{we}\mathrm{have}\mid \frac{2}{3}\mid <1\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}=0$$$$\mathrm{b})\mathrm{\Sigma }{\mathrm{u}}_{\mathrm{n}}\mathrm{with}{\mathrm{u}}_{\mathrm{n}}=\frac{{(-1)}^{\mathrm{n}}{\mathrm{n}}^2}{{3\mathrm{n}}^4+{2\mathrm{n}}^3+5\mathrm{n}-2}\Rightarrow \mid {\mathrm{u}}_{\mathrm{n}}\mid \sim \frac{{\mathrm{n}}^2}{{3\mathrm{n}}^4}=\frac{1}{{3\mathrm{n}}^2}$$$$\Rightarrow \mathrm{\Sigma }{\mathrm{u}}_{\mathrm{n}}\mathrm{converge}\mathrm{absolument}$$$$\mathrm{c})\mathrm{\Sigma }{\mathrm{v}}_{\mathrm{n}}\mathrm{with}{\mathrm{v}}_{\mathrm{n}}=\left(5\mathrm{n}\right)!{(\mathrm{x}-2)}^{\mathrm{n}}\Rightarrow$$$$\left({}^{\mathrm{n}}\sqrt{\mid {\mathrm{v}}_{\mathrm{n}}\mid }\right)={(\left(5\mathrm{n}\right)!)}^{\frac{1}{\mathrm{n}}}\mid \mathrm{x}-2\mid$$$${(\left(5\mathrm{n}\right)!)}^{\frac{1}{\mathrm{n}}}={\mathrm{e}}^{\frac{1}{\mathrm{n}}\log \left(5\mathrm{n}\right)!}\mathrm{we}\mathrm{have}\mathrm{n}!\sim {\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}\Rightarrow$$$$)5\mathrm{n})!\sim {\left(5\mathrm{n}\right)}^{5\mathrm{n}}{\mathrm{e}}^{-5\mathrm{n}}\sqrt{10\pi \mathrm{n}}\Rightarrow$$$$\log (5\mathrm{n}!)\sim 5\mathrm{nlog}\left(5\mathrm{n}\right)-5\mathrm{n}+\frac{1}{2}\log \left(10\pi \mathrm{n}\right)\Rightarrow$$$$\frac{\log (5\mathrm{n}!)}{\mathrm{n}}\sim 5\log \left(5\mathrm{n}\right)-5+\frac{1}{2\mathrm{n}}\log \left(10\pi \mathrm{n}\right)\to +\mathrm{\infty }\mathrm{so}\mathrm{for}\mathrm{x}\ne 2$$$${(\mid {\mathrm{v}}_{\mathrm{n}}\mid )}^{\frac{1}{\mathrm{n}}}\to \mathrm{\infty }\Rightarrow \mathrm{\Sigma }{\mathrm{v}}_{\mathrm{n}}\mathrm{diverges}...!$$

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