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Question Number 40040 by abdo mathsup 649 cc last updated on 15/Jul/18
![let A_n = ∫_0 ^n e^(−n( x+2−[x])) dx with n integr natural 1) calculate A_n 2) find lim_(n→+∞) A_n 3) study the convergence of Σ_n A_n](Q40040.png)
$${let}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:{e}^{−{n}\left(\:{x}+\mathrm{2}−\left[{x}\right]\right)} {dx}\:\:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{study}\:{the}\:{convergence}\:{of}\:\:\:\sum_{{n}} {A}_{{n}} \\ $$
Commented by abdo mathsup 649 cc last updated on 17/Jul/18
![1) we have A_n = Σ_(k=0) ^(n−1) ∫_k ^(k+1) e^(−n(x+2−k)) dx = Σ_(k=0) ^(n−1) e^(−n(2−k)) ∫_k ^(k+1) e^(−nx) dx =Σ_(k=0) ^(n−1) e^(n(k−2)) [−(1/n) e^(−nx) ]_k ^(k+1) = (1/n)Σ_(k=0) ^(n−1) e^(n(k−2)) { e^(−nk) −e^(−(n+1)k) } =(1/n) Σ_(k=0) ^(n−1) e^(−2n) −(1/n) Σ_(k=0) ^(n−1) e^(−2n−k) = e^(−2n) −(e^(−2n) /n) Σ_(k=0) ^(n−1) (e^(−1) )^k A_n =e^(−2n) −(e^(−2n) /n) ((1−(e^(−1) )^n )/(1−e^(−1) )) 2) we have lim_(n→+∞) e^(−2n) =0 lim_(n→+∞) (e^(−2n) /n) =0 and lim_(n→+∞) ((1− e^(−n) )/(2−e^(−1) )) =(1/(1−e^(−1) )) ⇒ lim_(n→+∞) A_n =0 3) we have Σ_(n=1) ^∞ A_n = Σ_(n=1) ^∞ e^(−2n) −(1/(1−e^(−1) ))Σ_(n=1) ^∞ (e^(−2n) /n) + (1/(1−e^(−1) )) Σ_(n=1) ^∞ (e^(−3n) /n) but we have Σ_(n=1) ^∞ e^(−2n) =Σ_(n=1) ^∞ (e^(−2) )^n =(1/(1−e^(−2) )) Σ_(n=1) ^∞ (e^(−2n) /n) =Σ_(n=1) ^∞ (((e^(−2) )^n )/n) =−ln(1−e^(−2) ) Σ_(n=1) ^∞ (e^(−3n) /n) =−ln(1−e^(−3) ) ⇒Σ A_n is convergent and Σ_(n≥1) A_n = (1/(1−e^(−2) )) +(1/(1−e^(−1) ))ln(1−e^(−2) ) −(1/(1−e^(−1) ))ln(1−e^(−3) ) .](Q40235.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\:{A}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:{e}^{−{n}\left({x}+\mathrm{2}−{k}\right)} {dx} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\:{e}^{−{n}\left(\mathrm{2}−{k}\right)} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:{e}^{−{nx}} {dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{e}^{{n}\left({k}−\mathrm{2}\right)} \:\:\left[−\frac{\mathrm{1}}{{n}}\:{e}^{−{nx}} \right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{e}^{{n}\left({k}−\mathrm{2}\right)} \left\{\:{e}^{−{nk}} \:−{e}^{−\left({n}+\mathrm{1}\right){k}} \right\} \\ $$$$=\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{e}^{−\mathrm{2}{n}} \:\:\:−\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:{e}^{−\mathrm{2}{n}−{k}} \\ $$$$=\:{e}^{−\mathrm{2}{n}} \:\:\:−\frac{{e}^{−\mathrm{2}{n}} }{{n}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({e}^{−\mathrm{1}} \right)^{{k}} \\ $$$${A}_{{n}} ={e}^{−\mathrm{2}{n}} \:\:−\frac{{e}^{−\mathrm{2}{n}} }{{n}}\:\frac{\mathrm{1}−\left({e}^{−\mathrm{1}} \right)^{{n}} }{\mathrm{1}−{e}^{−\mathrm{1}} } \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{lim}_{{n}\rightarrow+\infty} \:{e}^{−\mathrm{2}{n}} \:=\mathrm{0} \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:\frac{{e}^{−\mathrm{2}{n}} }{{n}}\:=\mathrm{0}\:\:{and}\:{lim}_{{n}\rightarrow+\infty} \:\:\frac{\mathrm{1}−\:{e}^{−{n}} }{\mathrm{2}−{e}^{−\mathrm{1}} }\:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} } \\ $$$$\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:\:{A}_{{n}} =\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:{we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{A}_{{n}} =\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−\mathrm{2}{n}} \:−\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} }\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{−\mathrm{2}{n}} }{{n}} \\ $$$$+\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} }\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{e}^{−\mathrm{3}{n}} }{{n}}\:{but}\:{we}\:{have}\: \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−\mathrm{2}{n}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left({e}^{−\mathrm{2}} \right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{2}} } \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{e}^{−\mathrm{2}{n}} }{{n}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left({e}^{−\mathrm{2}} \right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{1}−{e}^{−\mathrm{2}} \right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{e}^{−\mathrm{3}{n}} }{{n}}\:=−{ln}\left(\mathrm{1}−{e}^{−\mathrm{3}} \right)\:\Rightarrow\Sigma\:{A}_{{n}} \:{is}\:{convergent} \\ $$$${and}\:\sum_{{n}\geqslant\mathrm{1}} \:{A}_{{n}} =\:\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} }{ln}\left(\mathrm{1}−{e}^{−\mathrm{2}} \right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} }{ln}\left(\mathrm{1}−{e}^{−\mathrm{3}} \right)\:. \\ $$$$ \\ $$