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Question Number 74498 by mathmax by abdo last updated on 25/Nov/19
![1) calculte A_n =∫_0 ^∞ e^(−nx) [e^x ] dx with n integr and n≥2 2)find lim_(n→+∞) n^n A_n](Q74498.png)
$$\left.\mathrm{1}\right)\:{calculte}\:\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−{nx}} \left[{e}^{{x}} \right]\:{dx}\:\:\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{2} \\ $$$$\left.\mathrm{2}\right){find}\:{lim}_{{n}\rightarrow+\infty} \:{n}^{{n}} \:{A}_{{n}} \\ $$
Commented by mathmax by abdo last updated on 25/Nov/19
![1) A_n =∫_0 ^∞ e^(−nx) [e^x ]dx changement e^x =t give A_n =∫_1 ^(+∞) e^(−nlnt) [t](dt/t) =∫_1 ^(+∞) (([t])/t^(n+1) )dt =Σ_(k=1) ^∞ ∫_k ^(k+1) (k/t^(n+1) )dt =Σ_(k=1) ^∞ k ∫_k ^(k+) t^(−n−1) dt =Σ_(k=1) ^∞ k[−(1/n)t^(−n) ]_k ^(k+1) =Σ_(k=1) ^∞ (k/n){(1/k^n )−(1/k^(n+1) )} =(1/n)Σ_(k=1) ^∞ (1/k^(n−1) )−(1/n)Σ_(k=1) ^∞ (1/k^n ) we have ξ(x)=Σ_(k=1) ^∞ (1/k^x ) (x>1) ⇒A_n =(1/n)ξ(n−1)−(1/n)ξ(n) 2) we have n^n A_n =n^(n−1) {ξ(n−1)−ξ(n)} ⇒n^n A_n ∼c n^(n−1) ⇒ lim_(n→+∞) n^n A_n =∞](Q74512.png)
$$\left.\mathrm{1}\right)\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{nx}} \left[{e}^{{x}} \right]{dx}\:\:{changement}\:{e}^{{x}} ={t}\:\:{give} \\ $$$${A}_{{n}} =\int_{\mathrm{1}} ^{+\infty} \:{e}^{−{nlnt}} \left[{t}\right]\frac{{dt}}{{t}}\:=\int_{\mathrm{1}} ^{+\infty} \frac{\left[{t}\right]}{{t}^{{n}+\mathrm{1}} }{dt}\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\frac{{k}}{{t}^{{n}+\mathrm{1}} }{dt} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\infty} {k}\:\int_{{k}} ^{{k}+} \:\:{t}^{−{n}−\mathrm{1}} {dt}\:=\sum_{{k}=\mathrm{1}} ^{\infty} {k}\left[−\frac{\mathrm{1}}{{n}}{t}^{−{n}} \right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\infty} \frac{{k}}{{n}}\left\{\frac{\mathrm{1}}{{k}^{{n}} }−\frac{\mathrm{1}}{{k}^{{n}+\mathrm{1}} }\right\}\:=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{{n}−\mathrm{1}} }−\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{{n}} } \\ $$$${we}\:{have}\:\xi\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}^{{x}} }\:\:\:\left({x}>\mathrm{1}\right)\:\Rightarrow{A}_{{n}} =\frac{\mathrm{1}}{{n}}\xi\left({n}−\mathrm{1}\right)−\frac{\mathrm{1}}{{n}}\xi\left({n}\right) \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{n}^{{n}} {A}_{{n}} ={n}^{{n}−\mathrm{1}} \left\{\xi\left({n}−\mathrm{1}\right)−\xi\left({n}\right)\right\}\:\Rightarrow{n}^{{n}} \:{A}_{{n}} \sim{c}\:{n}^{{n}−\mathrm{1}} \:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{n}^{{n}} \:{A}_{{n}} =\infty \\ $$
Answered by mind is power last updated on 25/Nov/19
![A_n =Σ_(k≥1) ∫_(ln(k)) ^(ln(k+1)) e^(−nx) [e^x ]dx =Σ_(k≥1) ∫_(ln(k)) ^(ln(k+1)) .ke^(−nx) dx =Σ_(k≥1) ((−k)/n)[e^(−nx) ]_(ln(k)) ^(ln(k+1)) =Σ_(k≥1) ((−k)/(n(k+1)^n ))+(k/(nk^n )) =Σ_(k≥1) {((−1)/(n(k+1)^(n−1) ))+(1/(nk^(n−1) ))+(1/(n(k+1)^n ))} =(1/n)+Σ_(k≥1) (1/(n(k+1)^n ))=A_n n^n An≥n^(n−1) →+∞](Q74505.png)
$${A}_{{n}} =\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{{ln}\left({k}\right)} ^{{ln}\left({k}+\mathrm{1}\right)} {e}^{−{nx}} \left[{e}^{{x}} \right]{dx} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{{ln}\left({k}\right)} ^{{ln}\left({k}+\mathrm{1}\right)} .{ke}^{−{nx}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{−{k}}{{n}}\left[{e}^{−{nx}} \right]_{{ln}\left({k}\right)} ^{{ln}\left({k}+\mathrm{1}\right)} =\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{−{k}}{{n}\left({k}+\mathrm{1}\right)^{{n}} }+\frac{{k}}{{nk}^{{n}} } \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\left\{\frac{−\mathrm{1}}{{n}\left({k}+\mathrm{1}\right)^{{n}−\mathrm{1}} }+\frac{\mathrm{1}}{{nk}^{{n}−\mathrm{1}} }+\frac{\mathrm{1}}{{n}\left({k}+\mathrm{1}\right)^{{n}} }\right\} \\ $$$$=\frac{\mathrm{1}}{{n}}+\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}\left({k}+\mathrm{1}\right)^{{n}} }={A}_{{n}} \\ $$$${n}^{{n}} {An}\geqslant{n}^{{n}−\mathrm{1}} \rightarrow+\infty \\ $$$$ \\ $$