Question Number 41410 by maxmathsup by imad last updated on 06/Aug/18
![let A_n =∫_0 ^∞ [ne^(−x) ]dx with n≥2 1) calculate A_n 2) find nature of Σ_(n≥2) A_n 3) study the convergence of Σ (1/A_n ) and Σ (1/A_n ^2 )](https://www.tinkutara.com/question/Q41410.png)
$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\left[{ne}^{−{x}} \right]{dx}\:\:{with}\:{n}\geqslant\mathrm{2} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nature}\:{of}\:\sum_{{n}\geqslant\mathrm{2}} \:\:\:{A}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{study}\:{the}\:{convergence}\:{of}\:\:\Sigma\:\frac{\mathrm{1}}{{A}_{{n}} }\:\:{and}\:\Sigma\:\frac{\mathrm{1}}{{A}_{{n}} ^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 07/Aug/18
![1) changement n e^(−x) =t give e^(−x) =(t/n) ⇒e^x =(n/t) ⇒x =ln(n)−ln(t) ⇒ dx =−(dt/t) ⇒ A_n =− ∫_n ^0 [t] (dt/t) = ∫_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−1) k{ln(k+1)−ln(k)}⇒ A_n =Σ_(k=1) ^(n−1) k ln(1+(1/k))](https://www.tinkutara.com/question/Q41473.png)
$$\left.\mathrm{1}\right)\:{changement}\:{n}\:{e}^{−{x}} \:={t}\:{give}\:{e}^{−{x}} \:=\frac{{t}}{{n}}\:\Rightarrow{e}^{{x}} \:=\frac{{n}}{{t}}\:\Rightarrow{x}\:={ln}\left({n}\right)−{ln}\left({t}\right)\:\Rightarrow \\ $$$${dx}\:=−\frac{{dt}}{{t}}\:\Rightarrow\:{A}_{{n}} =−\:\int_{{n}} ^{\mathrm{0}} \:\:\:\left[{t}\right]\:\:\frac{{dt}}{{t}}\:=\:\int_{\mathrm{0}} ^{{n}} \:\:\frac{\left[{t}\right]}{{t}}\:{dt}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\frac{{k}}{{t}}\:{dt} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}\:\left\{{ln}\left({k}+\mathrm{1}\right)−{ln}\left({k}\right)\right\}\:=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} {k}\left\{{ln}\left({k}+\mathrm{1}\right)−{ln}\left({k}\right)\right\}\Rightarrow \\ $$$${A}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{k}\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right) \\ $$
Commented by maxmathsup by imad last updated on 07/Aug/18
![2) we have ln^′ (1+x) =(1/(1+x))=1−x +x^2 −... if ∣x∣<1 ⇒ ln(1+x) =x−(x^2 /2) +(x^3 /3) −... ⇒ ln(1+x)≥x−(x^2 /2) ⇒ln(1+(1/(k )))≥(1/k) −(1/(2k^2 )) ⇒ ∀ k∈[[0,n−1]] kln(1+(1/k))≥1 −(1/(2k)) ⇒Σ_(k=1) ^(n−1) k ln(1+(1/k)) ≥Σ_(k=1) ^(n−1) (1−(1/(2k))) ⇒ A_n ≥ n−1 −(1/2) Σ_(k=1) ^(n−1) (1/k) ⇒ A_n ≥ n−1 −(1/2) H_(n−1) but H_(n−1) =ln(n−1) +γ +o((1/n)) ⇒n−1−(1/2) H_(n−1) =n−1−ln(√(n−1)) −(γ/2) +o((1/n)) but lim_(n→+∞) n−1−ln(√(n−1)) =+∞ ⇒ A_n →+∞ ⇒ Σ A_n diverges](https://www.tinkutara.com/question/Q41474.png)
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}}=\mathrm{1}−{x}\:\:+{x}^{\mathrm{2}} \:−…\:{if}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}\right)\:={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:−…\:\Rightarrow\:{ln}\left(\mathrm{1}+{x}\right)\geqslant{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}\:}\right)\geqslant\frac{\mathrm{1}}{{k}}\:−\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\forall\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:\:{kln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)\geqslant\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{2}{k}}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} {k}\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)\:\geqslant\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)\:\Rightarrow \\ $$$${A}_{{n}} \geqslant\:\:{n}−\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:\Rightarrow\:{A}_{{n}} \geqslant\:{n}−\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}−\mathrm{1}} \:\:\:{but} \\ $$$${H}_{{n}−\mathrm{1}} ={ln}\left({n}−\mathrm{1}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow{n}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}−\mathrm{1}} ={n}−\mathrm{1}−{ln}\sqrt{{n}−\mathrm{1}}\:−\frac{\gamma}{\mathrm{2}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$${but}\:{lim}_{{n}\rightarrow+\infty} {n}−\mathrm{1}−{ln}\sqrt{{n}−\mathrm{1}}\:=+\infty\:\Rightarrow\:{A}_{{n}} \:\rightarrow+\infty\:\Rightarrow\:\Sigma\:{A}_{{n}} \:{diverges} \\ $$