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Question Number 106246 by mohammad17 last updated on 03/Aug/20

Commented by mohammad17 last updated on 03/Aug/20

$$testtheseriesconvergeordiverge$$

Commented by mathmax by abdo last updated on 04/Aug/20

$$9)\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}\mathrm{with}{\mathrm{u}}_{\mathrm{n}}=\frac{{(\mathrm{n}+5)}^2}{{(\mathrm{n}+3)}^5}=\phi \left(\mathrm{n}\right)\mathrm{with}$$$$\phi \left(\mathrm{x}\right)=\frac{{(\mathrm{x}+5)}^2}{{(\mathrm{x}+3)}^5}(\mathrm{we}\mathrm{take}\mathrm{x}>0)\Rightarrow$$$${\phi }^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{2(\mathrm{x}+5){(\mathrm{x}+3)}^5-5{(\mathrm{x}+3)}^4{(\mathrm{x}+5)}^2}{{(\mathrm{x}+3)}^{10}}$$$$=\frac{2(\mathrm{x}+5)(\mathrm{x}+3)-5{(\mathrm{x}+5)}^2}{{(\mathrm{x}+3)}^6}=\frac{{2\mathrm{x}}^2+6\mathrm{x}+10\mathrm{x}+30-5({\mathrm{x}}^2+10\mathrm{x}+25)}{{(\mathrm{x}+3)}^6}$$$$=\frac{{2\mathrm{x}}^2+16\mathrm{x}+30-{5\mathrm{x}}^2-50\mathrm{x}-125}{{(\mathrm{x}+3)}^6}$$$$=\frac{-{3\mathrm{x}}^2-34\mathrm{x}-125}{{(\mathrm{x}+3)}^2}<0\Rightarrow \phi \mathrm{is}\mathrm{decreazing}\Rightarrow \mathrm{S}\mathrm{is}\mathrm{alternate}\mathrm{serie}$$$$\mathrm{convergent}.$$$$$$

Commented by mohammad17 last updated on 04/Aug/20

$$thankyousir$$

Answered by mathmax by abdo last updated on 03/Aug/20

$$3)\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+2^{\mathrm{n}}}\mathrm{for}\mathrm{all}\mathrm{n}⩾1\mathrm{n}+2^{\mathrm{n}}>2^{\mathrm{n}}\Rightarrow \frac{1}{\mathrm{n}+2^{\mathrm{n}}}<\frac{1}{2^{\mathrm{n}}}\Rightarrow$$$$\mid \frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+2^{\mathrm{n}}}\mid <\frac{1}{2^{\mathrm{n}}}\mathrm{and}\mathrm{the}\mathrm{serie}\mathrm{\Sigma }\frac{1}{2^{\mathrm{n}}}\mathrm{converges}\Rightarrow \mathrm{S}\mathrm{converves}$$

Commented by mohammad17 last updated on 04/Aug/20

$$thankyousir$$

Commented by mathmax by abdo last updated on 04/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}$$

Answered by mathmax by abdo last updated on 03/Aug/20

$$5)\mathrm{S}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{{\left(\mathrm{nln}\left(\mathrm{n}\right)\right)}^2}\Rightarrow \mathrm{S}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}-1}{\mathrm{u}}_{\mathrm{n}}$$$${\mathrm{u}}_{\mathrm{n}}=\frac{1}{{\left(\mathrm{nln}\left(\mathrm{n}\right)\right)}^2}{\mathrm{u}}_{\mathrm{n}}\mathrm{is}>0\mathrm{decrezing}\mathrm{to}0\Rightarrow \mathrm{S}\mathrm{is}\mathrm{a}\mathrm{alternate}\mathrm{serie}$$$$\mathrm{convergente}$$$$\mathrm{another}\mathrm{way}\mid \frac{{(-1)}^{\mathrm{n}-1}}{{\left(\mathrm{nlnn}\right)}^2}\mid ⩽\frac{1}{{\left(\mathrm{nlnn}\right)}^2}\mathrm{the}\mathrm{sdquence}{\mathrm{u}}_{\mathrm{n}}=\frac{1}{{\left(\mathrm{nlnn}\right)}^2}$$$$\mathrm{is}\mathrm{decreazing}\Rightarrow \sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{1}{{\left(\mathrm{nlnn}\right)}^2}\mathrm{and}{\int }_2^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\left(\mathrm{xlnx}\right)}^2}\mathrm{have}\mathrm{same}$$$$\mathrm{nature}\mathrm{of}\mathrm{convergence}\mathrm{we}\mathrm{have}$$$${\int }_2^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\left(\mathrm{xlnx}\right)}^2}{=}_{\mathrm{lnx}=\mathrm{t}}{\int }_{\ln 2}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}}{{\left({\mathrm{e}}^{\mathrm{t}}\mathrm{t}\right)}^2}={\int }_{\ln 2}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}}{{\mathrm{t}}^2}\mathrm{dt}\mathrm{and}\mathrm{this}\mathrm{integral}\mathrm{is}$$$$\mathrm{convergent}\Rightarrow \mathrm{S}\mathrm{is}\mathrm{convergent}$$

Commented by mohammad17 last updated on 04/Aug/20

$$thankyousir$$

Commented by mathmax by abdo last updated on 04/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}$$

Answered by mathmax by abdo last updated on 04/Aug/20

$$7)\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\sin \left(\alpha \mathrm{n}\right)}{\mathrm{n}!}\Rightarrow \mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}$$$$\mid \frac{{\mathrm{u}}_{\mathrm{n}+1}}{{\mathrm{u}}_{\mathrm{n}}}\mid =\mid \frac{\sin \left(\alpha (\mathrm{n}+1)\right)}{(\mathrm{n}+1)!}\times \frac{\mathrm{n}!}{\sin \left(\alpha \mathrm{n}\right)}\mid =\mid \frac{\sin (\alpha \mathrm{n}+\mathrm{n})}{\sin \left(\alpha \mathrm{n}\right)}\mid \times \frac{1}{\mathrm{n}+1}\to 0(\mathrm{n}\to +\mathrm{\infty })$$$$\mathrm{S}\mathrm{is}\mathrm{convervent}$$$$\mathrm{another}\mathrm{way}\mathrm{we}\mathrm{have}\mid \frac{\sin \left(\alpha \mathrm{n}\right)}{\mathrm{n}!}\mid ⩽\frac{1}{\mathrm{n}!}\mathrm{\forall }\mathrm{n}\mathrm{and}\mathrm{\Sigma }\frac{1}{\mathrm{n}!}\mathrm{converges}\Rightarrow$$$$\mathrm{S}\mathrm{converges}\mathrm{absolument}\Rightarrow \mathrm{S}\mathrm{conerges}..$$

Commented by mohammad17 last updated on 04/Aug/20

$$thankyousir$$

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