n-1-log-n-n-2-Does-the-series-converge-or-diverge-help-find-the-sum-

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

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

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

let ϕ(x)=(((ln(x))/x))^2 with x≥2 we have ϕ^′ (x)=2 ((ln(x))/x)(((ln(x))/x))^′ =((2lnx)/x){((1−ln(x))/x^2 )} <0 ϕ is decreasing on [2,+∞[ so Σ_(n=1) ^∞ ( ((ln(n))/n))^2 and ∫_2 ^(+∞) (((ln(x))/x))^2 dx have the same nature of convergence but ∫_2 ^(+∞) (((ln(x))^2 )/x^2 ) dx =_(ln(x)=t) ∫_(ln(2)) ^()∞) (t^2 /e^(2t) ) e^t dt = ∫_(ln(2)) ^(+∞) t^2 e^(−t) dt and this integral converges ⇒ Σ_(n=1) ^∞ (((ln(n))/n))^2 converges.

$${let}\:\varphi\left({x}\right)=\left(\frac{{ln}\left({x}\right)}{{x}}\right)^{\mathrm{2}} \:{with}\:{x}\geqslant\mathrm{2}\:{we}\:{have}\:\varphi^{'} \left({x}\right)=\mathrm{2}\:\frac{{ln}\left({x}\right)}{{x}}\left(\frac{{ln}\left({x}\right)}{{x}}\right)^{'} \\ $$$$=\frac{\mathrm{2}{lnx}}{{x}}\left\{\frac{\mathrm{1}−{ln}\left({x}\right)}{{x}^{\mathrm{2}} }\right\}\:<\mathrm{0}\:\varphi\:{is}\:{decreasing}\:{on}\:\left[\mathrm{2},+\infty\left[\:{so}\:\sum_{{n}=\mathrm{1}} ^{\infty} \left(\:\frac{{ln}\left({n}\right)}{{n}}\right)^{\mathrm{2}} \:{and}\right.\right. \\ $$$$\int_{\mathrm{2}} ^{+\infty} \:\left(\frac{{ln}\left({x}\right)}{{x}}\right)^{\mathrm{2}} {dx}\:{have}\:{the}\:{same}\:{nature}\:{of}\:{convergence}\:{but} \\ $$$$\int_{\mathrm{2}} ^{+\infty} \:\frac{\left({ln}\left({x}\right)\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:{dx}\:=_{{ln}\left({x}\right)={t}} \:\:\:\int_{{ln}\left(\mathrm{2}\right)} ^{\left.\right)\infty} \:\:\:\frac{{t}^{\mathrm{2}} }{{e}^{\mathrm{2}{t}} }\:\:{e}^{{t}} \:{dt}\:=\:\int_{{ln}\left(\mathrm{2}\right)} ^{+\infty} \:{t}^{\mathrm{2}} \:{e}^{−{t}} {dt}\:{and}\:{this}\:{integral}\: \\ $$$${converges}\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(\frac{{ln}\left({n}\right)}{{n}}\right)^{\mathrm{2}} \:{converges}. \\ $$

Commented by MJS last updated on 28/Oct/18

it seems that ∫_1 ^(+∞) (((ln x)/x))^2 dx=2

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

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

Please how can i prove it is monotonic first sir

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

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

$${thank}\:{you}\:{sir}. \\ $$

Facebook Tweet Pin

n-1-log-n-n-2-Does-the-series-converge-or-diverge-help-find-the-sum-

Leave a Reply Cancel reply