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Question Number 140825 by liberty last updated on 13/May/21

$$\mathrm{Use}\mathrm{the}\mathrm{limit}\mathrm{comparison}\mathrm{test}$$$$\mathrm{to}\mathrm{determine}\mathrm{if}\mathrm{the}\mathrm{series}\mathrm{converges}$$$$\mathrm{or}\mathrm{diverges}$$$$\underset{\mathrm{n}=2}{\sum ^{\mathrm{\infty }}}\frac{1}{7+8\mathrm{n}\ln \left(\ln \mathrm{n}\right)}.$$

Answered by mathmax by abdo last updated on 13/May/21

$$\mathrm{let}\phi \left(\mathrm{x}\right)=\frac{1}{7+8\mathrm{nln}\left(\mathrm{lnn}\right)}(\mathrm{n}>2)\Rightarrow {\phi }^{{}^{\prime }}\left(\mathrm{x}\right)=-\frac{{(7+8\mathrm{xlog}\left(\mathrm{logx}\right))}^{{}^{\prime }}}{(7+8\mathrm{xlog}{\left(\mathrm{logx}\right)}^2}$$$$=-\frac{8\log \left(\mathrm{logx}\right)+8\mathrm{x}\times \frac{1}{\mathrm{xlogx}}}{{(....)}^2}=-\frac{8\log \left(\mathrm{logx}\right)+\frac{8}{\mathrm{logx}}}{{(....)}^2}<0\Rightarrow \phi \mathrm{is}\mathrm{decreazing}$$$$\mathrm{what}\mathrm{about}{\int }_{\mathrm{e}}^{\mathrm{\infty }}\frac{\mathrm{dx}}{7+8\mathrm{x}\log \left(\mathrm{logx}\right)}?\mathrm{changement}\mathrm{logx}=\mathrm{t}\mathrm{give}$$$${\int }_{\mathrm{e}}^{\mathrm{\infty }}\frac{\mathrm{dx}}{7+8\mathrm{xlog}\left(\mathrm{logx}\right)}={\int }_1^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}}{7+{8\mathrm{e}}^{\mathrm{t}}\log \left(\mathrm{t}\right)}={\int }_1^{\mathrm{\infty }}\frac{\mathrm{dt}}{{7\mathrm{e}}^{-\mathrm{t}}+8\mathrm{logt}}$$$$\mathrm{at}+\mathrm{\infty }\frac{1}{{7\mathrm{e}}^{-\mathrm{t}}+8\mathrm{logt}}\sim \frac{1}{8\mathrm{logt}}\mathrm{and}$$$${\int }_1^{\mathrm{\infty }}\frac{\mathrm{dt}}{\mathrm{logt}}{=}_{\mathrm{logt}=\mathrm{u}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{u}}}{\mathrm{u}}\mathrm{du}{\lim }_{\mathrm{u}\to +\mathrm{\infty }}\sqrt{\mathrm{u}}.\frac{{\mathrm{e}}^{\mathrm{u}}}{\mathrm{u}}=+\mathrm{\infty }\Rightarrow \mathrm{this}\mathrm{integral}\mathrm{diverges}$$$$\Rightarrow \mathrm{\Sigma }\frac{1}{7+8\mathrm{nlog}\left(\mathrm{logn}\right)}\mathrm{diverges}...!$$

Commented by liberty last updated on 13/May/21

$$\mathrm{why}\mathrm{sir}{\int }_{\mathrm{e}}^{\mathrm{\infty }}\frac{\mathrm{dx}}{7+8\mathrm{x}\ln \left(\ln \mathrm{x}\right)}?$$$$\mathrm{it}\mathrm{should}\mathrm{be}{\int }_2^{\mathrm{\infty }}...\mathrm{dx}\mathrm{sir}?$$

Commented by Mathspace last updated on 13/May/21

$${\int }_2^{\mathrm{\infty }}...dxand{\int }_e^{+\mathrm{\infty }}...dxarethesame$$$$forconvergence(notvalue!)$$

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