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Question Number 38101 by maxmathsup by imad last updated on 21/Jun/18

let  A_n = ∫_0 ^n   e^(x−[x]) dx  1) calculate A_n   2) find lim_(n→+∞)  A_n

$${let}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:{e}^{{x}−\left[{x}\right]} {dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$

Commented by prof Abdo imad last updated on 22/Jun/18

A_n =Σ_(k=0) ^(n−1)  ∫_k ^(k+1)  e^(−k)  e^x dx  =Σ_(k=0) ^(n−1)  e^(−k)   ( e^(k+1)  −e^k )  =Σ_(k=0) ^(n−1) (e −1)=(e−1)Σ_(k=0) ^(n−1) (1)=n(e−1)  A_n =n(e−1)  2)we have e−1>0 ⇒ lim_(n→+∞)  A_n =+∞.

$${A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{e}^{−{k}} \:{e}^{{x}} {dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{−{k}} \:\:\left(\:{e}^{{k}+\mathrm{1}} \:−{e}^{{k}} \right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({e}\:−\mathrm{1}\right)=\left({e}−\mathrm{1}\right)\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{1}\right)={n}\left({e}−\mathrm{1}\right) \\ $$$${A}_{{n}} ={n}\left({e}−\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:{e}−\mathrm{1}>\mathrm{0}\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =+\infty. \\ $$

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