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Question Number 133568 by bemath last updated on 23/Feb/21

$$\mathrm{Proof}\mathrm{the}\mathrm{series}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{2}{9+2\mathrm{n}{\left(\ln \mathrm{n}\right)}^2}$$$$\mathrm{convergent}$$$$$$

Answered by EDWIN88 last updated on 23/Feb/21

$$\mathrm{let}{\mathrm{b}}_{\mathrm{n}}=\frac{2}{9+2\mathrm{n}{\left(\ln \mathrm{n}\right)}^2}\mathrm{and}{\mathrm{a}}_{\mathrm{n}}=\frac{2}{2\mathrm{n}{\left(\ln \mathrm{n}\right)}^2}$$$$\mathrm{we}\mathrm{know}\mathrm{that}\frac{2}{9+2\mathrm{n}{\left(\ln \mathrm{n}\right)}^2}⩽\frac{2}{2\mathrm{n}{\left(\ln \mathrm{n}\right)}^2}=\frac{1}{\mathrm{n}{\left(\ln \mathrm{n}\right)}^2}$$$$\mathrm{then}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{2}{9+2\mathrm{n}{\left(\ln \mathrm{n}\right)}^2}⩽\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}{\left(\ln \mathrm{n}\right)}^2}$$$$$$$$\mathrm{consider}\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}{\left(\ln \mathrm{n}\right)}^2}=0,\mathrm{it}\mathrm{follows}\mathrm{that}$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}{\left(\ln \mathrm{n}\right)}^2}\mathrm{convergent},\mathrm{then}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{2}{9+2\mathrm{n}{\left(\ln \mathrm{n}\right)}^2}$$$$\mathrm{also}\mathrm{convergent}$$$$$$

Answered by mathmax by abdo last updated on 24/Feb/21

$$\frac{2}{9+2\mathrm{n}{\left(\mathrm{lnn}\right)}^2}⩽\frac{1}{\mathrm{n}(\ln {\left(\mathrm{n}\right)}^2}\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{2}{9+2\mathrm{n}{\left(\mathrm{lnn}\right)}^2}⩽\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{1}{\mathrm{n}{\left(\mathrm{lnn}\right)}^2}$$$$\mathrm{the}\mathrm{serie}{\mathrm{u}}_{\mathrm{n}}=\frac{1}{\mathrm{n}{\left(\mathrm{lnn}\right)}^2}\mathrm{is}\mathrm{decreazing}\mathrm{to}\mathrm{o}\mathrm{so}\mathrm{its}\mathrm{nature}\mathrm{is}\mathrm{same}\mathrm{to}$$$${\int }_2^{\mathrm{\infty }}\frac{\mathrm{dx}}{\mathrm{x}{\left(\mathrm{lnx}\right)}^2}\mathrm{and}\mathrm{changement}\mathrm{lnx}=\mathrm{t}\mathrm{give}$$$${\int }_2^{\mathrm{\infty }}\frac{\mathrm{dx}}{\mathrm{x}{\left(\mathrm{lnx}\right)}^2}={\int }_{\ln 2}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}}{{\mathrm{e}}^{\mathrm{t}}.{\mathrm{t}}^2}={\int }_{\ln 2}^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2}={[-\frac{1}{\mathrm{t}}]}_{\ln 2}^{\mathrm{\infty }}=\frac{1}{\ln 2}<+\mathrm{\infty }\Rightarrow \mathrm{this}\mathrm{serie}$$$$\mathrm{is}\mathrm{convergent}...!$$

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