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Question Number 46527 by Tawa1 last updated on 28/Oct/18

$$\underset{\mathrm{n}=1}{\mathbf{\sum }^{\mathrm{\infty }}}{\left(\frac{\log \mathrm{n}}{\mathrm{n}}\right)}^2$$$$\mathrm{Does}\mathrm{the}\mathrm{series}\mathrm{converge}\mathrm{or}\mathrm{diverge},\mathrm{help}\mathrm{find}\mathrm{the}\mathrm{sum}...$$

Commented by maxmathsup by imad last updated on 28/Oct/18

$$let\phi \left(x\right)={\left(\frac{ln\left(x\right)}{x}\right)}^{2}withx⩾2wehave{\phi }^{{}^{\prime }}\left(x\right)=2\frac{ln\left(x\right)}{x}{\left(\frac{ln\left(x\right)}{x}\right)}^{{}^{\prime }}$$$$=\frac{2lnx}{x}\left\{\frac{1-ln\left(x\right)}{x^2}\right\}<0\phi isdecreasingon[2,+\mathrm{\infty }[so\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{ln\left(n\right)}{n}\right)}^{2}and$$$${\int }_2^{+\mathrm{\infty }}{\left(\frac{ln\left(x\right)}{x}\right)}^{2}dxhavethesamenatureofconvergencebut$$$${\int }_2^{+\mathrm{\infty }}\frac{{\left(ln\left(x\right)\right)}^2}{x^2}dx{=}_{ln\left(x\right)=t}{\int }_{ln\left(2\right)}^{)\mathrm{\infty }}\frac{t^2}{e^{2t}}{e}^{t}dt={\int }_{ln\left(2\right)}^{+\mathrm{\infty }}{t}^2{e}^{-t}dtandthisintegral$$$$converges\Rightarrow \sum _{n=1}^{\mathrm{\infty }}{\left(\frac{ln\left(n\right)}{n}\right)}^{2}converges.$$

Commented by MJS last updated on 28/Oct/18

$$\mathrm{it}\mathrm{seems}\mathrm{that}\underset{1}{\stackrel{+\mathrm{\infty }}{\int }}{\left(\frac{\ln x}{x}\right)}^{2}dx=2$$

Commented by Tawa1 last updated on 28/Oct/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 28/Oct/18

$$\mathrm{Please}\mathrm{how}\mathrm{can}\mathrm{i}\mathrm{prove}\mathrm{it}\mathrm{is}\mathrm{monotonic}\mathrm{first}\mathrm{sir}$$

Commented by math khazana by abdo last updated on 28/Oct/18

$$thankyousir.$$

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