Question Number 43170 by maxmathsup by imad last updated on 07/Sep/18
![let A_n = ∫_0 ^∞ [n e^(−x) ]dx with n integr natural. 1) calculate A_n 2) find lim_(n→+∞) A_n .](Q43170.png)
$${let}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\left[{n}\:{e}^{−{x}} \right]{dx}\:\:{with}\:{n}\:{integr}\:{natural}. \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} . \\ $$
Commented by alex041103 last updated on 08/Sep/18
![is [] the whole part function e.g. [2.67]=2???](Q43239.png)
$${is}\:\left[\right]\:{the}\:{whole}\:{part}\:{function} \\ $$$${e}.{g}.\:\left[\mathrm{2}.\mathrm{67}\right]=\mathrm{2}??? \\ $$
Commented by maxmathsup by imad last updated on 08/Sep/18
![look sir alex for all x from R x=[x] +r with 0≤r<1 and if x ∈Z [x]=x we have 2,67 =2 +0,67 and 0,67 ∈[0,1[ ⇒[2,67]=2 .also we have [x]≤x<[x] +1 .](Q43243.png)
$${look}\:\:{sir}\:{alex}\:{for}\:{all}\:{x}\:{from}\:{R}\:\:{x}=\left[{x}\right]\:+{r}\:{with}\:\:\mathrm{0}\leqslant{r}<\mathrm{1}\:{and}\:{if}\:{x}\:\in{Z} \\ $$$$\left[{x}\right]={x}\:\:\:{we}\:{have}\:\mathrm{2},\mathrm{67}\:=\mathrm{2}\:+\mathrm{0},\mathrm{67}\:\:{and}\:\mathrm{0},\mathrm{67}\:\in\left[\mathrm{0},\mathrm{1}\left[\:\Rightarrow\left[\mathrm{2},\mathrm{67}\right]=\mathrm{2}\:.{also}\:{we}\:{have}\right.\right. \\ $$$$\left[{x}\right]\leqslant{x}<\left[{x}\right]\:+\mathrm{1}\:. \\ $$
Commented by maxmathsup by imad last updated on 09/Sep/18
) = ∫_0 ^n (([t])/t)dt = Σ_(k=0) ^(n−1) ∫_k ^(k+1) (k/t)dt =Σ_(k=0) ^(n−1) k {ln(k+1)−ln(k)} = Σ_(k=1) ^n k{ln(k+1)−ln(k)} ⇒A_n = Σ_(k=1) ^n k ln(((k+1)/k)) . 2) we have A_n =Σ_(k=1) ^n k ln(1+(1/k)) but ln(1+x) = x−(x^2 /2) +o(x^3 ) x−(x^2 /2) ≤ln(1+x)≤x ⇒(1/k)−(1/(2k^2 ))≤ln(1+(1/k))≤(1/k) ⇒ k ln(1+(1/k))≥1− (1/(2k)) ⇒ A_n ≥ Σ_(k=1) ^n (1−(1/(2k))) ⇒A_n ≥ n−(1/2) H_n but H_n = ln(n)+γ +o((1/n)) ⇒n−(1/2) H_n =n−(1/2)ln(n)−(γ/2) +o((1/n)) but lim_(n→+∞) n−((ln(n))/2) =lim_(n→+∞) n(1−((ln(n))/(2n)))=+∞ ⇒lim_(n→+∞) A_n =+∞ .](Q43329.png)
$$\left.\mathrm{1}\right)\:{changement}\:{ne}^{−{x}} ={t}\:{give}\:{e}^{−{x}} =\frac{{t}}{{n}}\:\:\Rightarrow−{x}\:={ln}\left({t}\right)−{ln}\left({n}\right)\:\Rightarrow−{dx}=\frac{{dt}}{{t}} \\ $$$${and}\:{A}_{{n}} =\:−\int_{\mathrm{0}} ^{{n}} \:\left[{t}\right]\left(−\frac{{dt}}{{t}}\right)\:=\:\int_{\mathrm{0}} ^{{n}} \:\:\:\frac{\left[{t}\right]}{{t}}{dt}\:=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\frac{{k}}{{t}}{dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}\:\left\{{ln}\left({k}+\mathrm{1}\right)−{ln}\left({k}\right)\right\}\:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\left\{{ln}\left({k}+\mathrm{1}\right)−{ln}\left({k}\right)\right\} \\ $$$$\Rightarrow{A}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{ln}\left(\frac{{k}+\mathrm{1}}{{k}}\right)\:. \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:\:{A}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)\:\:\:\:{but}\:\: \\ $$$${ln}\left(\mathrm{1}+{x}\right)\:=\:{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right)\:\:{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\leqslant{ln}\left(\mathrm{1}+{x}\right)\leqslant{x}\:\Rightarrow\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} }\leqslant{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)\leqslant\frac{\mathrm{1}}{{k}}\:\Rightarrow \\ $$$${k}\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)\geqslant\mathrm{1}−\:\frac{\mathrm{1}}{\mathrm{2}{k}}\:\Rightarrow\:{A}_{{n}} \:\geqslant\:\sum_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)\:\Rightarrow{A}_{{n}} \geqslant\:{n}−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \\ $$$${but}\:{H}_{{n}} =\:{ln}\left({n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow{n}−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} ={n}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}\right)−\frac{\gamma}{\mathrm{2}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:{but} \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:{n}−\frac{{ln}\left({n}\right)}{\mathrm{2}}\:={lim}_{{n}\rightarrow+\infty} \:{n}\left(\mathrm{1}−\frac{{ln}\left({n}\right)}{\mathrm{2}{n}}\right)=+\infty\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {A}_{{n}} =+\infty\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18
![∫_0 ^(log_e n) [ne^(−x) ]dx+∫_(log_e n) ^∞ [ne^(−x) ]dx now ne^(−x) =1 e^(−x) =n^(−1) e^x =n x=log_e n when x=log_e n ne^(−x) =1 ∫_0 ^(log_e n) 1×dx+∫_0 ^∞ [ne^(−x) ]dx =∣x∣_0 ^(log_e n) +0 =log_e n](Q43325.png)
$$\int_{\mathrm{0}} ^{{log}_{{e}} {n}} \left[{ne}^{−{x}} \right]{dx}+\int_{{log}_{{e}} {n}} ^{\infty} \left[{ne}^{−{x}} \right]{dx} \\ $$$${now}\:{ne}^{−{x}} =\mathrm{1} \\ $$$${e}^{−{x}} ={n}^{−\mathrm{1}} \\ $$$${e}^{{x}} ={n} \\ $$$${x}={log}_{{e}} {n}\:\:{when}\:{x}={log}_{{e}} {n}\:\:{ne}^{−{x}} =\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{{log}_{{e}} {n}} \mathrm{1}×{dx}+\int_{\mathrm{0}} ^{\infty} \left[{ne}^{−{x}} \right]{dx} \\ $$$$=\mid{x}\mid_{\mathrm{0}} ^{{log}_{{e}} {n}} +\mathrm{0} \\ $$$$={log}_{{e}} {n} \\ $$