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Use-the-limit-comparison-test-to-determine-if-the-series-converges-or-diverges-n-2-1-7-8n-ln-ln-n-

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Question Number 140825 by liberty last updated on 13/May/21
Use the limit comparison test to determine if the series converges or diverges Σ_(n=2) ^∞ (1/(7+8n ln (ln n))).
$$\mathrm{Use}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{comparison}\:\mathrm{test} \\ $$$$\mathrm{to}\:\mathrm{determine}\:\mathrm{if}\:\mathrm{the}\:\mathrm{series}\:\mathrm{converges} \\ $$$$\mathrm{or}\:\mathrm{diverges}\: \\ $$$$\:\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{7}+\mathrm{8n}\:\mathrm{ln}\:\left(\mathrm{ln}\:\mathrm{n}\right)}.\: \\ $$
Answered by mathmax by abdo last updated on 13/May/21
let ϕ(x)=(1/(7+8nln(lnn))) (n>2) ⇒ϕ^′ (x)=−(((7+8xlog(logx))^′ )/((7+8xlog(logx)^2 )) =−((8log(logx) +8x×(1/(xlogx)))/((....)^2 )) =−((8log(logx)+(8/(logx)))/((....)^2 ))<0 ⇒ ϕ is decreazing what about ∫_e ^∞ (dx/(7+8x log(logx))) ?changement logx=t give ∫_e ^∞ (dx/(7+8xlog(logx))) =∫_1 ^∞ ((e^t dt)/(7+8e^t log(t))) =∫_1 ^∞ (dt/(7e^(−t) +8logt)) at +∞ (1/(7e^(−t) +8logt))∼(1/(8logt)) and ∫_1 ^∞ (dt/(logt)) =_(logt=u) ∫_0 ^∞ (e^u /u)du lim_(u→+∞) (√u).(e^u /u)=+∞ ⇒this integral diverges ⇒Σ (1/(7+8nlog(logn))) diverges...!
$$\mathrm{let}\:\varphi\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{7}+\mathrm{8nln}\left(\mathrm{lnn}\right)}\:\:\left(\mathrm{n}>\mathrm{2}\right)\:\Rightarrow\varphi^{'} \left(\mathrm{x}\right)=−\frac{\left(\mathrm{7}+\mathrm{8xlog}\left(\mathrm{logx}\right)\right)^{'} }{\left(\mathrm{7}+\mathrm{8xlog}\left(\mathrm{logx}\right)^{\mathrm{2}} \right.} \\ $$$$=−\frac{\mathrm{8log}\left(\mathrm{logx}\right)\:+\mathrm{8x}×\frac{\mathrm{1}}{\mathrm{xlogx}}}{\left(….\right)^{\mathrm{2}} }\:=−\frac{\mathrm{8log}\left(\mathrm{logx}\right)+\frac{\mathrm{8}}{\mathrm{logx}}}{\left(….\right)^{\mathrm{2}} }<\mathrm{0}\:\Rightarrow\:\varphi\:\mathrm{is}\:\mathrm{decreazing} \\ $$$$\mathrm{what}\:\mathrm{about}\:\int_{\mathrm{e}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{7}+\mathrm{8x}\:\mathrm{log}\left(\mathrm{logx}\right)}\:\:?\mathrm{changement}\:\mathrm{logx}=\mathrm{t}\:\mathrm{give} \\ $$$$\int_{\mathrm{e}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{7}+\mathrm{8xlog}\left(\mathrm{logx}\right)}\:=\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{e}^{\mathrm{t}} \mathrm{dt}}{\mathrm{7}+\mathrm{8e}^{\mathrm{t}} \mathrm{log}\left(\mathrm{t}\right)}\:=\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dt}}{\mathrm{7e}^{−\mathrm{t}} \:+\mathrm{8logt}} \\ $$$$\mathrm{at}\:+\infty\:\:\:\frac{\mathrm{1}}{\mathrm{7e}^{−\mathrm{t}} \:+\mathrm{8logt}}\sim\frac{\mathrm{1}}{\mathrm{8logt}}\:\mathrm{and} \\ $$$$\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dt}}{\mathrm{logt}}\:=_{\mathrm{logt}=\mathrm{u}} \:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{\mathrm{u}} }{\mathrm{u}}\mathrm{du}\:\:\:\:\:\mathrm{lim}_{\mathrm{u}\rightarrow+\infty} \sqrt{\mathrm{u}}.\frac{\mathrm{e}^{\mathrm{u}} }{\mathrm{u}}=+\infty\:\Rightarrow\mathrm{this}\:\mathrm{integral}\:\mathrm{diverges} \\ $$$$\Rightarrow\Sigma\:\frac{\mathrm{1}}{\mathrm{7}+\mathrm{8nlog}\left(\mathrm{logn}\right)}\:\mathrm{diverges}…! \\ $$
Commented by liberty last updated on 13/May/21
why sir ∫_e ^∞ (dx/(7+8x ln (ln x))) ? it should be ∫_2 ^∞ ... dx sir?
$$\mathrm{why}\:\mathrm{sir}\:\int_{\mathrm{e}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{7}+\mathrm{8x}\:\mathrm{ln}\:\left(\mathrm{ln}\:\mathrm{x}\right)}\:? \\ $$$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\int_{\mathrm{2}} ^{\infty} …\:\mathrm{dx}\:\mathrm{sir}? \\ $$
Commented by Mathspace last updated on 13/May/21
∫_2 ^∞ ...dx and ∫_e ^(+∞) ...dx are the same for convergence ( not value !)
$$\int_{\mathrm{2}} ^{\infty} \:\:…{dx}\:{and}\:\int_{{e}} ^{+\infty} …{dx}\:{are}\:{the}\:{same} \\ $$$${for}\:{convergence}\:\left(\:{not}\:{value}\:!\right) \\ $$

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