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Question Number 44473 by abdo.msup.com last updated on 29/Sep/18

let A_n =∫_0 ^∞  sin(n[t])e^(−t) dt  2)calculate A_n   and lim_(n→+∞) n A_n   3)study the convergence of Σ_n  A_n

$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{sin}\left({n}\left[{t}\right]\right){e}^{−{t}} {dt} \\ $$$$\left.\mathrm{2}\right){calculate}\:{A}_{{n}} \:\:{and}\:{lim}_{{n}\rightarrow+\infty} {n}\:{A}_{{n}} \\ $$$$\left.\mathrm{3}\right){study}\:{the}\:{convergence}\:{of}\:\sum_{{n}} \:{A}_{{n}} \\ $$

Commented by maxmathsup by imad last updated on 30/Sep/18

1) we have A_n  = Σ_(k=0) ^∞   ∫_k ^(k+1)  sin(nk)e^(−t) dt  =Σ_(k=0) ^∞  sin(kn) ∫_k ^(k+1)  e^(−t) dt =Σ_(k=0) ^∞  sin(kn)[−e^(−t) ]_k ^(k+1)   =Σ_(k=0) ^∞   sin(kn){ e^(−k)  −e^(−(k+1)) }=Σ_(k=0) ^∞  e^(−k)  sin(kn)−Σ_(k=0) ^∞  e^(−(k+1))  sin(kn)  but Σ_(k=0) ^∞  e^(−k)  sin(kn)=Im( Σ_(k=0) ^∞  e^(−k+ikn) )  and  Σ_(k=0) ^∞   e^(−k +ikn)  =Σ_(k=0) ^∞  e^((−1+in)k) =(1/(1−e^(−1+in) )) =(1/(1−e^(−1) (cos(n)+isin(n))))  = (1/(1−e^(−1) cosn−ie^(−1) sin(n))) =((1−e^(−1) cos(n) +ie^(−1) sin(n))/((1−e^(−1) cos(n))^2  +e^(−2) sin^2 n)) ⇒  Σ_(k=0) ^∞  e^(−k) sin(kn) =((e^(−1) sin(n))/((1−e^(−1) cosn)^2  +e^(−2) sin^2 n)) also  Σ_(k=0) ^∞  e^(−(k+1)) sin(kn) =e^(−1)  Σ_(k=0) ^∞  e^(−k)  sin(kn) =((e^(−2) sin(n))/((1−e^(−1) cos(n))^2  +e^(−2) sin^2 n))  A_n =(1−e^(−1) ) ((e^(−1) sin(n))/(1−2e^(−1) cosn +e^(−2) )) =(((e^(−1)  −d^(−2) )sin(n))/(1−2 e^(−1) cos(n) +e^(−2) ))  A_n =(((e−1)sin(n))/(e^2  −2e cos(n) +1)) .

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}_{{n}} \:=\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:{sin}\left({nk}\right){e}^{−{t}} {dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \:{sin}\left({kn}\right)\:\int_{{k}} ^{{k}+\mathrm{1}} \:{e}^{−{t}} {dt}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:{sin}\left({kn}\right)\left[−{e}^{−{t}} \right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \:\:{sin}\left({kn}\right)\left\{\:{e}^{−{k}} \:−{e}^{−\left({k}+\mathrm{1}\right)} \right\}=\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{k}} \:{sin}\left({kn}\right)−\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−\left({k}+\mathrm{1}\right)} \:{sin}\left({kn}\right) \\ $$$${but}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{k}} \:{sin}\left({kn}\right)={Im}\left(\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{k}+{ikn}} \right)\:\:{and} \\ $$$$\sum_{{k}=\mathrm{0}} ^{\infty} \:\:{e}^{−{k}\:+{ikn}} \:=\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{\left(−\mathrm{1}+{in}\right){k}} =\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}+{in}} }\:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} \left({cos}\left({n}\right)+{isin}\left({n}\right)\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} {cosn}−{ie}^{−\mathrm{1}} {sin}\left({n}\right)}\:=\frac{\mathrm{1}−{e}^{−\mathrm{1}} {cos}\left({n}\right)\:+{ie}^{−\mathrm{1}} {sin}\left({n}\right)}{\left(\mathrm{1}−{e}^{−\mathrm{1}} {cos}\left({n}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}} {sin}^{\mathrm{2}} {n}}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{k}} {sin}\left({kn}\right)\:=\frac{{e}^{−\mathrm{1}} {sin}\left({n}\right)}{\left(\mathrm{1}−{e}^{−\mathrm{1}} {cosn}\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}} {sin}^{\mathrm{2}} {n}}\:{also} \\ $$$$\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−\left({k}+\mathrm{1}\right)} {sin}\left({kn}\right)\:={e}^{−\mathrm{1}} \:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{k}} \:{sin}\left({kn}\right)\:=\frac{{e}^{−\mathrm{2}} {sin}\left({n}\right)}{\left(\mathrm{1}−{e}^{−\mathrm{1}} {cos}\left({n}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}} {sin}^{\mathrm{2}} {n}} \\ $$$${A}_{{n}} =\left(\mathrm{1}−{e}^{−\mathrm{1}} \right)\:\frac{{e}^{−\mathrm{1}} {sin}\left({n}\right)}{\mathrm{1}−\mathrm{2}{e}^{−\mathrm{1}} {cosn}\:+{e}^{−\mathrm{2}} }\:=\frac{\left({e}^{−\mathrm{1}} \:−{d}^{−\mathrm{2}} \right){sin}\left({n}\right)}{\mathrm{1}−\mathrm{2}\:{e}^{−\mathrm{1}} {cos}\left({n}\right)\:+{e}^{−\mathrm{2}} } \\ $$$${A}_{{n}} =\frac{\left({e}−\mathrm{1}\right){sin}\left({n}\right)}{{e}^{\mathrm{2}} \:−\mathrm{2}{e}\:{cos}\left({n}\right)\:+\mathrm{1}}\:. \\ $$

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