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Question Number 113897 by Lordose last updated on 16/Sep/20

Answered by mathmax by abdo last updated on 17/Sep/20

$$\mathrm{let}{\mathrm{u}}_{\mathrm{n}}=\frac{3^{\mathrm{n}}}{\sqrt{(3\mathrm{n}-2)2^{\mathrm{n}}}}\Rightarrow {\mathrm{u}}_{\mathrm{n}}=\frac{{\left(\frac{3}{\sqrt{2}}\right)}^{\mathrm{n}}}{\sqrt{3\mathrm{n}-2}}(\mathrm{n}⩾1)\Rightarrow$$$$\frac{{\mathrm{u}}_{\mathrm{n}+1}}{{\mathrm{u}}_{\mathrm{n}}}=\frac{{\left(\frac{3}{\sqrt{2}}\right)}^{\mathrm{n}+1}}{\sqrt{3\mathrm{n}+1}}\times \frac{\sqrt{3\mathrm{n}-2}}{{\left(\frac{3}{\sqrt{2}}\right)}^{\mathrm{n}}}=\sqrt{\frac{3\mathrm{n}-2}{3\mathrm{n}+1}}\times \left(\frac{3}{\sqrt{2}}\right)\mathrm{and}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{{\mathrm{u}}_{\mathrm{n}+1}}{{\mathrm{u}}_{\mathrm{n}}}$$$$=\frac{3}{\sqrt{2}}\mathrm{so}\mathrm{for}\mid \mathrm{x}\mid <\frac{3}{\sqrt{2}}\mathrm{the}\mathrm{serie}\mathrm{converges}\mathrm{and}\mathrm{for}\mid \mathrm{x}\mid ⩾\frac{3}{\sqrt{2}}$$$$\mathrm{the}\mathrm{serie}\mathrm{diverges}\mathrm{finally}{\mathrm{I}}_{\mathrm{c}}=]-\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}[$$

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