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Question Number 33717 by prof Abdo imad last updated on 22/Apr/18
find the value of Σ_(n=0) ^∞ artan( (((√(n+1)) −(√n))/(1+(√(n^2 +n)))) )
$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{artan}\left(\:\frac{\sqrt{{n}+\mathrm{1}}\:−\sqrt{{n}}}{\mathrm{1}+\sqrt{{n}^{\mathrm{2}} +{n}}}\:\right) \\ $$
Commented by prof Abdo imad last updated on 24/Apr/18
let put S_n = Σ_(k=0) ^n arctan((((√(k+1)) −(√k))/(1+(√(k^2 +k))))) let put (√k)=tan(u_k ) ⇔u_k =arctan((√k)) ⇒ S_n = Σ_(k=0) ^n arctan( ((tan(u_(k+1) ) −tan(u_k ))/(1+tan(u_k )tan(u_(k+1) )))) =Σ_(k=0) ^n arctan(tan(u_(k+1) −u_k )) =Σ_(k=0) ^n (u_(k+1) −u_k ) =u_(n+1) −u_0 = arctan((√n)) ⇒ lim_(n→+∞) S_n = (π/2) .
$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{arctan}\left(\frac{\sqrt{{k}+\mathrm{1}}\:−\sqrt{{k}}}{\mathrm{1}+\sqrt{{k}^{\mathrm{2}} +{k}}}\right)\:{let}\:{put} \\ $$$$\sqrt{{k}}={tan}\left({u}_{{k}} \right)\:\Leftrightarrow{u}_{{k}} ={arctan}\left(\sqrt{{k}}\right)\:\Rightarrow \\ $$$${S}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{arctan}\left(\:\frac{{tan}\left({u}_{{k}+\mathrm{1}} \right)\:−{tan}\left({u}_{{k}} \right)}{\mathrm{1}+{tan}\left({u}_{{k}} \right){tan}\left({u}_{{k}+\mathrm{1}} \right)}\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{arctan}\left({tan}\left({u}_{{k}+\mathrm{1}} −{u}_{{k}} \right)\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \left({u}_{{k}+\mathrm{1}} \:−{u}_{{k}} \right)\:={u}_{{n}+\mathrm{1}} \:−{u}_{\mathrm{0}} =\:{arctan}\left(\sqrt{{n}}\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\:\frac{\pi}{\mathrm{2}}\:. \\ $$
Commented by prof Abdo imad last updated on 24/Apr/18
S_n =arctan((√(n+1)))
$${S}_{{n}} ={arctan}\left(\sqrt{{n}+\mathrm{1}}\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Apr/18
Σ_0 ^∞ tan^(−1) ((√(n+1))) −tan^(−1) ((√(n))) T_(n ) =tan^(−1) ((√(n+1 ))) −tan^(−1) ((√(n))) T_0 =tan^(−1) ((√(1))) −tan^(−1) (0) T_1 =tan^(−1) ((√(2))) −tan^(−1) (1) so S_n =T_0 +T_1 +.....+T_n =tan^(−1) ((√(n+1)) )−tan^(−1) (0) when n tends to infinity the sum is tan^(−1) (∞) −tan^(−1) (0) =∐/2 −0 =∐/2 ∐ this sign used as pie
$$\underset{\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\left.{n}+\mathrm{1}\right)}\:−\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\left.{n}\right)}\right.\right. \\ $$$${T}_{{n}\:} =\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\left.{n}+\mathrm{1}\:\right)}\:\:−\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\left.{n}\right)}\right.\right. \\ $$$${T}_{\mathrm{0}} =\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\left.\mathrm{1}\right)}\:−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{0}\right)\right. \\ $$$${T}_{\mathrm{1}} =\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\left.\mathrm{2}\right)}\:−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)\right. \\ $$$${so}\:{S}_{{n}} ={T}_{\mathrm{0}} \:+{T}_{\mathrm{1}} +…..+{T}_{{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{{n}+\mathrm{1}}\:\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{0}\right) \\ $$$${when}\:{n}\:{tends}\:{to}\:{infinity}\:{the}\:{sum}\:{is} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\infty\right)\:−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{0}\right) \\ $$$$=\coprod/\mathrm{2}\:−\mathrm{0} \\ $$$$=\coprod/\mathrm{2} \\ $$$$\coprod\:{this}\:{sign}\:{used}\:{as}\:{pie}\: \\ $$$$ \\ $$$$ \\ $$

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