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Question Number 145165 by mathmax by abdo last updated on 02/Jul/21

calculate Σ_(n=0) ^∞ arctan(((2n+1)/(n^4 +2n^3 +n^2 +1)))

$$\mathrm{calculate}\:\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \mathrm{arctan}\left(\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{n}^{\mathrm{4}} \:+\mathrm{2n}^{\mathrm{3}} \:+\mathrm{n}^{\mathrm{2}} \:+\mathrm{1}}\right) \\ $$

Answered by mnjuly1970 last updated on 02/Jul/21

Ω_n = Σ_(k=0) ^n arctan((((k+1)^2 − k^^2 )/(k^( 2) ( k+1 )^( 2) +1)) ) =Σ_(k=0) ^n {arctan (k+1)^2 − arctan(k^( 2) )} =^(Telescopic series) arctan( n +1 )^( 2) −arctan (0 ) Ans : = lim_( n→∞) Ω_( n) = arctan (∞ )−arctan( 0 )=(π/2)

$$\:\:\:\:\:\:\Omega_{{n}} =\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:{arctan}\left(\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\:{k}^{\:^{\mathrm{2}} } }{{k}^{\:\mathrm{2}} \left(\:{k}+\mathrm{1}\:\right)^{\:\mathrm{2}} \:+\mathrm{1}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\left\{{arctan}\:\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\:{arctan}\left({k}^{\:\mathrm{2}} \right)\right\} \\ $$$$\:\:\:\overset{\mathrm{Telescopic}\:\mathrm{series}} {=}\:{arctan}\left(\:{n}\:+\mathrm{1}\:\right)^{\:\mathrm{2}} −{arctan}\:\left(\mathrm{0}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Ans}\::\:=\:\mathrm{lim}_{\:{n}\rightarrow\infty} \:\Omega_{\:{n}} \:=\:{arctan}\:\left(\infty\:\right)−{arctan}\left(\:\mathrm{0}\:\right)=\frac{\pi}{\mathrm{2}} \\ $$

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