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Which-of-the-series-converge-and-which-diverge-Check-by-the-limit-comparison-test-1-n-2-1-n-ln-n-n-2-5-2-n-1-ln-n-n-3-2-3-n-3-1-ln-lnn-4-n-1-

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Question Number 67244 by Learner-123 last updated on 24/Aug/19
Which of the series converge and which diverge? Check by the limit comparison test. 1) Σ_(n=2) ^∞ ((1+n ln(n))/(n^2 +5)) 2) Σ_(n=1) ^∞ ((ln(n))/n^(3/2) ) 3) Σ_(n=3) ^∞ (1/(ln(lnn))) 4) Σ_(n=1) ^∞ (1/(n (n)^(1/n) )) ??
$${Which}\:{of}\:{the}\:{series}\:{converge}\:{and}\: \\ $$$${which}\:{diverge}?\:{Check}\:{by}\:{the}\:{limit} \\ $$$${comparison}\:{test}. \\ $$$$\left.\mathrm{1}\right)\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}+{n}\:{ln}\left({n}\right)}{{n}^{\mathrm{2}} +\mathrm{5}} \\ $$$$\left.\mathrm{2}\right)\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{ln}\left({n}\right)}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\left.\mathrm{3}\right)\:\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{ln}\left({lnn}\right)} \\ $$$$\left.\mathrm{4}\right)\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\:\left({n}\right)^{\frac{\mathrm{1}}{{n}}} }\:\: \\ $$$$?? \\ $$
Commented by mathmax by abdo last updated on 24/Aug/19
1)Σ_(n=2) ^∞ ((1+nln(n))/(n^2 +5)) =Σ_(n=2) ^∞ (1/(n^2 +5)) +Σ_(n=2) ^∞ ((nln(n))/(n^2 +5)) (1/(n^2 +5)) =(1/(n^2 (1+5n^(−2) )))∼(1/n^2 ) so this serie converges ((nln(n))/(n^2 +5)) =((nln(n))/(n^2 (1+5n^(−2) ))) ∼ ((ln(n))/n) let determine nature of Σ((ln(n))/n) f(t)=((ln(t))/t) with t≥2 ⇒f^′ (t) =((1−lnt)/t^2 )<0 ⇒f decreases ⇒ Σ((ln(n))/n) and ∫_2 ^(+∞) ((ln(t))/t)dt are the same nature changement ln(t)=u give ∫_2 ^(+∞) ((ln(t))/t)dt =∫_(ln(2)) ^(+∞) (u/e^u ) e^u du =∫_(ln(2)) ^(+∞) udu =+∞ so this serie diverges.
$$\left.\mathrm{1}\right)\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}+{nln}\left({n}\right)}{{n}^{\mathrm{2}} \:+\mathrm{5}}\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{5}}\:+\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{nln}\left({n}\right)}{{n}^{\mathrm{2}} \:+\mathrm{5}} \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{5}}\:=\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{5}{n}^{−\mathrm{2}} \right)}\sim\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:{so}\:{this}\:{serie}\:{converges} \\ $$$$\frac{{nln}\left({n}\right)}{{n}^{\mathrm{2}} \:+\mathrm{5}}\:=\frac{{nln}\left({n}\right)}{{n}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{5}{n}^{−\mathrm{2}} \right)}\:\sim\:\frac{{ln}\left({n}\right)}{{n}}\:\:{let}\:{determine}\:{nature}\:{of}\:\Sigma\frac{{ln}\left({n}\right)}{{n}} \\ $$$${f}\left({t}\right)=\frac{{ln}\left({t}\right)}{{t}}\:\:{with}\:{t}\geqslant\mathrm{2}\:\Rightarrow{f}^{'} \left({t}\right)\:=\frac{\mathrm{1}−{lnt}}{{t}^{\mathrm{2}} }<\mathrm{0}\:\Rightarrow{f}\:\:{decreases}\:\:\Rightarrow \\ $$$$\Sigma\frac{{ln}\left({n}\right)}{{n}}\:\:{and}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{{ln}\left({t}\right)}{{t}}{dt}\:\:{are}\:{the}\:{same}\:{nature} \\ $$$$\:{changement}\:{ln}\left({t}\right)={u}\:\:{give}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{{ln}\left({t}\right)}{{t}}{dt}\:=\int_{{ln}\left(\mathrm{2}\right)} ^{+\infty} \:\frac{{u}}{{e}^{{u}} }\:{e}^{{u}} {du}\: \\ $$$$=\int_{{ln}\left(\mathrm{2}\right)} ^{+\infty} \:{udu}\:=+\infty\:\:{so}\:{this}\:{serie}\:{diverges}. \\ $$
Commented by mathmax by abdo last updated on 24/Aug/19
2) let u_n =((ln(n))/n^(3/2) ) and f(x) =((lnx)/x^(3/2) ) with x>1 we have f^′ (x)=(((1/x)x^(3/2) −(3/2)x^(1/2) lnx)/x^3 ) =(((√x)−(3/2)(√x)ln(x))/x^3 ) =(1/x^(5/2) )×(1−lnx)<0 ⇒ f decrease so (u_n )decreases ⇒the serie has the nature of I∫_1 ^(+∞) ((ln(x))/x^(3/2) )dx changement ln(x)=t give I =∫_0 ^∞ (t/e^((3/2)t) )dt =∫_0 ^∞ t e^(−(3/2)t) dt its clear that I converges ⇒ this serie converges.
$$\left.\mathrm{2}\right)\:{let}\:{u}_{{n}} =\frac{{ln}\left({n}\right)}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:\:\:{and}\:{f}\left({x}\right)\:=\frac{{lnx}}{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:\:{with}\:{x}>\mathrm{1}\:\:{we}\:{have} \\ $$$${f}^{'} \left({x}\right)=\frac{\frac{\mathrm{1}}{{x}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{3}}{\mathrm{2}}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} {lnx}}{{x}^{\mathrm{3}} }\:=\frac{\sqrt{{x}}−\frac{\mathrm{3}}{\mathrm{2}}\sqrt{{x}}{ln}\left({x}\right)}{{x}^{\mathrm{3}} }\:=\frac{\mathrm{1}}{{x}^{\frac{\mathrm{5}}{\mathrm{2}}} }×\left(\mathrm{1}−{lnx}\right)<\mathrm{0}\:\Rightarrow \\ $$$${f}\:{decrease}\:{so}\:\left({u}_{{n}} \right){decreases}\:\:\Rightarrow{the}\:{serie}\:{has}\:{the}\:{nature}\:{of} \\ $$$${I}\int_{\mathrm{1}} ^{+\infty} \:\frac{{ln}\left({x}\right)}{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}\:\:\:{changement}\:{ln}\left({x}\right)={t}\:\:{give}\: \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}}{{e}^{\frac{\mathrm{3}}{\mathrm{2}}{t}} }{dt}\:=\int_{\mathrm{0}} ^{\infty} {t}\:{e}^{−\frac{\mathrm{3}}{\mathrm{2}}{t}} \:\:\:{dt}\:\:\:{its}\:{clear}\:{that}\:{I}\:{converges}\:\Rightarrow \\ $$$${this}\:{serie}\:{converges}. \\ $$
Commented by mathmax by abdo last updated on 24/Aug/19
f^′ (x) =(1/x^(5/2) )×(1−(3/2)ln(x))<0
$${f}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{{x}^{\frac{\mathrm{5}}{\mathrm{2}}} }×\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}}{ln}\left({x}\right)\right)<\mathrm{0} \\ $$
Commented by Learner-123 last updated on 25/Aug/19
Sir, in 1)→ f ′(t) = ((1−lnt)/t^2 ) <0 ∀ t ε (e,∞) . But ≥0 ∀ tε [2,e].
$$\left.{Sir},\:{in}\:\mathrm{1}\right)\rightarrow\:{f}\:'\left({t}\right)\:=\:\frac{\mathrm{1}−{lnt}}{{t}^{\mathrm{2}} }\:<\mathrm{0} \\ $$$$\:\forall\:{t}\:\epsilon\:\left({e},\infty\right)\:.\:{But}\:\geqslant\mathrm{0}\:\forall\:{t}\epsilon\:\left[\mathrm{2},{e}\right]. \\ $$
Commented by mathmax by abdo last updated on 25/Aug/19
sir the nature of a serie dont changes for a few terms...
$${sir}\:{the}\:{nature}\:{of}\:{a}\:{serie}\:{dont}\:\:{changes}\:{for}\:{a}\:{few}\:{terms}… \\ $$
Commented by Learner-123 last updated on 25/Aug/19
ok,thank you prof.
$${ok},{thank}\:{you}\:{prof}. \\ $$
Commented by mathmax by abdo last updated on 26/Aug/19
3) the sequence u_n =(1/(ln(lnn))) is decreasing on [3,+∞[ so Σ u_n and ∫_3 ^(+∞) (dx/(ln(lnx))) have the same nature changement lnx=t give ∫_3 ^(+∞) (dx/(ln(lnx))) =∫_(ln(3)) ^(+∞) ((e^t dt)/(ln(t))) we have lim_(t→+∞) (e^t /(lnt)) =+∞ ⇒ the integral diverges ⇒Σu_n diverges.
$$\left.\mathrm{3}\right)\:{the}\:{sequence}\:{u}_{{n}} =\frac{\mathrm{1}}{{ln}\left({lnn}\right)}\:{is}\:{decreasing}\:\:{on}\:\left[\mathrm{3},+\infty\left[\:{so}\right.\right. \\ $$$$\Sigma\:{u}_{{n}} \:\:{and}\:\:\int_{\mathrm{3}} ^{+\infty} \:\frac{{dx}}{{ln}\left({lnx}\right)}\:{have}\:{the}\:{same}\:{nature}\: \\ $$$${changement}\:{lnx}={t}\:{give}\:\int_{\mathrm{3}} ^{+\infty} \:\frac{{dx}}{{ln}\left({lnx}\right)}\:=\int_{{ln}\left(\mathrm{3}\right)} ^{+\infty} \:\frac{{e}^{{t}} \:{dt}}{{ln}\left({t}\right)} \\ $$$${we}\:{have}\:{lim}_{{t}\rightarrow+\infty} \:\:\:\:\frac{{e}^{{t}} }{{lnt}}\:\:=+\infty\:\Rightarrow\:{the}\:{integral}\:{diverges}\:\Rightarrow\Sigma{u}_{{n}} \\ $$$${diverges}. \\ $$

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