Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 68188 by mhmd last updated on 06/Sep/19

Σ_(n=3) ^∝  1/n(ln n)^2    is the function converg or diverg ? pleas help me

$$\sum_{{n}=\mathrm{3}} ^{\propto} \:\mathrm{1}/{n}\left({ln}\:{n}\right)^{\mathrm{2}} \:\:\:{is}\:{the}\:{function}\:{converg}\:{or}\:{diverg}\:?\:{pleas}\:{help}\:{me} \\ $$

Commented by Abdo msup. last updated on 07/Sep/19

let ϕ(x) =(1/(x(lnx)^2 ))  with  x≥3  we have  ϕ^′ (x) =−(((lnx)^2 +x ×((2lnx)/x))/(x^2 (lnx)^4 )) =−(((lnx)^2  ++2lnx)/(x^2 (lnx)^4 )) ⇒  ϕ^′ (x)<0 ⇒ϕ is increasing on [3,+∞[  so  Σ_(n=3) ^∞  (1/(n(lnn)^2 )) and ∫_3 ^(+∞)    (dx/(x(lnx)^2 )) have the same  nsture of convervence  changement lnx =t give  ∫_3 ^(+∞)   (dx/(x(lnx)^2 )) =∫_(ln(3)) ^(+∞)   ((e^t dt)/(e^t t^2 ))dt =∫_(ln(3)) ^(+∞)  (dt/t^2 )  =[−(1/t)]_(ln(3)) ^(+∞)  =(1/(ln(3)))<+∞ ⇒this integral converges  ⇒the serie  converges.

$${let}\:\varphi\left({x}\right)\:=\frac{\mathrm{1}}{{x}\left({lnx}\right)^{\mathrm{2}} }\:\:{with}\:\:{x}\geqslant\mathrm{3}\:\:{we}\:{have} \\ $$$$\varphi^{'} \left({x}\right)\:=−\frac{\left({lnx}\right)^{\mathrm{2}} +{x}\:×\frac{\mathrm{2}{lnx}}{{x}}}{{x}^{\mathrm{2}} \left({lnx}\right)^{\mathrm{4}} }\:=−\frac{\left({lnx}\right)^{\mathrm{2}} \:++\mathrm{2}{lnx}}{{x}^{\mathrm{2}} \left({lnx}\right)^{\mathrm{4}} }\:\Rightarrow \\ $$$$\varphi^{'} \left({x}\right)<\mathrm{0}\:\Rightarrow\varphi\:{is}\:{increasing}\:{on}\:\left[\mathrm{3},+\infty\left[\right.\right. \\ $$$${so}\:\:\sum_{{n}=\mathrm{3}} ^{\infty} \:\frac{\mathrm{1}}{{n}\left({lnn}\right)^{\mathrm{2}} }\:{and}\:\int_{\mathrm{3}} ^{+\infty} \:\:\:\frac{{dx}}{{x}\left({lnx}\right)^{\mathrm{2}} }\:{have}\:{the}\:{same} \\ $$$${nsture}\:{of}\:{convervence}\:\:{changement}\:{lnx}\:={t}\:{give} \\ $$$$\int_{\mathrm{3}} ^{+\infty} \:\:\frac{{dx}}{{x}\left({lnx}\right)^{\mathrm{2}} }\:=\int_{{ln}\left(\mathrm{3}\right)} ^{+\infty} \:\:\frac{{e}^{{t}} {dt}}{{e}^{{t}} {t}^{\mathrm{2}} }{dt}\:=\int_{{ln}\left(\mathrm{3}\right)} ^{+\infty} \:\frac{{dt}}{{t}^{\mathrm{2}} } \\ $$$$=\left[−\frac{\mathrm{1}}{{t}}\right]_{{ln}\left(\mathrm{3}\right)} ^{+\infty} \:=\frac{\mathrm{1}}{{ln}\left(\mathrm{3}\right)}<+\infty\:\Rightarrow{this}\:{integral}\:{converges} \\ $$$$\Rightarrow{the}\:{serie}\:\:{converges}. \\ $$

Commented by Abdo msup. last updated on 07/Sep/19

sorry ϕ is decreazing on[3,+∞[

$${sorry}\:\varphi\:{is}\:{decreazing}\:{on}\left[\mathrm{3},+\infty\left[\right.\right. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com