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Question Number 33716 by prof Abdo imad last updated on 22/Apr/18
calculate Σ_(n=0) ^∞ arctan( ((e^(n+1) −e^n )/(1+e^(2n+1) ))) .
$${calculate}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:{arctan}\left(\:\frac{{e}^{{n}+\mathrm{1}} \:\:−{e}^{{n}} }{\mathrm{1}+{e}^{\mathrm{2}{n}+\mathrm{1}} }\right)\:. \\ $$
Commented by prof Abdo imad last updated on 31/May/18
let put S_n = Σ_(k=0) ^n arctan( ((e^(n+1) −e^n )/(1+e^(2n+1) ))) let use the changement e^n =tan(u_n ) ⇒ u_n =arctan(e^n ) ((e^(n+1) −e^n )/(1+e^(2n+1) )) = ((tan(u_(n+1) ) −tan(u_n ))/(1+tan(u_(n+1) )tan(u_n ))) =tan(u_(n+1) −u_n ) ⇒ S_n =Σ_(k=0) ^n (u_(n+1) −u_n ) = u_(n+1) −u_0 = arctan(e^(n+1) ) −(π/4) ⇒lim_(n→+∞) S_n =(π/2) −(π/4) =(π/4) ★lim_(n→+∞) S_n = (π/4) ★
$${let}\:{put}\:{S}_{{n}} \:=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{arctan}\left(\:\frac{{e}^{{n}+\mathrm{1}} \:−{e}^{{n}} }{\mathrm{1}+{e}^{\mathrm{2}{n}+\mathrm{1}} }\right)\:{let}\:{use} \\ $$$${the}\:{changement}\:{e}^{{n}} \:={tan}\left({u}_{{n}} \right)\:\Rightarrow\:{u}_{{n}} ={arctan}\left({e}^{{n}} \right) \\ $$$$\frac{{e}^{{n}+\mathrm{1}} \:−{e}^{{n}} }{\mathrm{1}+{e}^{\mathrm{2}{n}+\mathrm{1}} }\:=\:\frac{{tan}\left({u}_{{n}+\mathrm{1}} \right)\:−{tan}\left({u}_{{n}} \right)}{\mathrm{1}+{tan}\left({u}_{{n}+\mathrm{1}} \right){tan}\left({u}_{{n}} \right)} \\ $$$$={tan}\left({u}_{{n}+\mathrm{1}} \:−{u}_{{n}} \right)\:\Rightarrow\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\left({u}_{{n}+\mathrm{1}} \:−{u}_{{n}} \right) \\ $$$$=\:{u}_{{n}+\mathrm{1}} \:−{u}_{\mathrm{0}} \:\:\:\:=\:{arctan}\left({e}^{{n}+\mathrm{1}} \right)\:−\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:=\frac{\pi}{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\:=\frac{\pi}{\mathrm{4}} \\ $$$$\bigstar{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:=\:\frac{\pi}{\mathrm{4}}\:\bigstar \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Apr/18
=Σ_(n=0) ^∞ {tan^(−1) (e^(n+1) )−tan^(−1) (e^n )} T_n =tan^(−1) (e^(n+1) )−tan^(−1) (e^n ) T_o =tan^(−1) (e^1 )−tan^(−1) (e^0 ) T_1 =tan^(−1) (e^2 )−tan^(−1) (e^1 ) T_2 =tan^(−1) (e^3 )−tan^(−1) (e^2 ) so adding upto n terms S_n =tan^(−1) (e^(n+1) )−tan^(−1) (e^o ) when n tends to infinity =tan^(−1) (∞)−tan^(−1) (1) =Π/2−Π/4 =Π/4
$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\mathrm{tan}^{−\mathrm{1}} \left({e}^{{n}+\mathrm{1}} \right)−\mathrm{tan}^{−\mathrm{1}} \left({e}^{{n}} \right)\right\} \\ $$$${T}_{{n}} =\mathrm{tan}^{−\mathrm{1}} \left({e}^{{n}+\mathrm{1}} \right)−\mathrm{tan}^{−\mathrm{1}} \left({e}^{{n}} \right) \\ $$$${T}_{{o}} =\mathrm{tan}^{−\mathrm{1}} \left({e}^{\mathrm{1}} \right)−\mathrm{tan}^{−\mathrm{1}} \left({e}^{\mathrm{0}} \right) \\ $$$$\:\:{T}_{\mathrm{1}} =\mathrm{tan}^{−\mathrm{1}} \left({e}^{\mathrm{2}} \right)−\mathrm{tan}^{−\mathrm{1}} \left({e}^{\mathrm{1}} \right) \\ $$$${T}_{\mathrm{2}} =\mathrm{tan}^{−\mathrm{1}} \left({e}^{\mathrm{3}} \right)−\mathrm{tan}^{−\mathrm{1}} \left({e}^{\mathrm{2}} \right) \\ $$$${so}\:{adding}\:{upto}\:{n}\:{terms}\: \\ $$$${S}_{{n}} =\mathrm{tan}^{−\mathrm{1}} \left({e}^{{n}+\mathrm{1}} \right)−\mathrm{tan}^{−\mathrm{1}} \left({e}^{{o}} \right) \\ $$$$\:\:{when}\:{n}\:{tends}\:{to}\:{infinity} \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\infty\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$=\Pi/\mathrm{2}−\Pi/\mathrm{4} \\ $$$$=\Pi/\mathrm{4} \\ $$

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