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Question Number 227051 by mnjuly1970 last updated on 28/Dec/25
Σ_(n∈N∪{0}) tan^(−1) ((1/(n^2 +3n + 2)) )=? ■
$$ \\ $$$$\:\:\:\:\:\underset{{n}\in\mathbb{N}\cup\left\{\mathrm{0}\right\}} {\sum}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}\:+\:\mathrm{2}}\:\right)=?\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Spillover last updated on 28/Dec/25
(π/4)?
$$\frac{\pi}{\mathrm{4}}? \\ $$
Commented by Ghisom_ last updated on 29/Dec/25
to everybody who answered: 1. Σ_(n=0) ^∞ T(n) =Σ_(n=0) ^4 T(n) +Σ_(n=5) ^∞ T(n) but Σ_(n=0) ^4 arctan (1/(n^2 +3n+2)) =arctan ((70293)/(68926)) >(π/4) and arctan (1/(n^2 +3n+2)) >0∀n∈N ⇒ Σ_(n=0) ^∞ arctan (1/(n^2 +3n+2)) >(π/4) 2. arctan u +arctan v =arctan ((u+v)/(1−uv)) [for some values of u, v] tan (α+β) =((tan α +tan β)/(1−tan α tan β)) ⇒ your answers are wrong changing the question to Σ_(n=0) ^∞ arctan (1/(n^2 +3n+3)) makes it easilysolvable because (1/(n^2 +3n+3))=(((n+2)−(n+1))/(1+(n+1)(n+2))) ⇒ arctan (1/(n^2 +3n+3)) =arctan (n+2) −arctan (n+1) but with (1/(n^2 +3n+2)) this doesn′t work
$$\mathrm{to}\:\mathrm{everybody}\:\mathrm{who}\:\mathrm{answered}: \\ $$$$\mathrm{1}. \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{T}\left({n}\right)\:=\underset{{n}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\:{T}\left({n}\right)\:+\underset{{n}=\mathrm{5}} {\overset{\infty} {\sum}}{T}\left({n}\right) \\ $$$$\mathrm{but} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\mathrm{arctan}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}\:=\mathrm{arctan}\:\frac{\mathrm{70293}}{\mathrm{68926}}\:>\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{and} \\ $$$$\mathrm{arctan}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}\:>\mathrm{0}\forall{n}\in\mathbb{N} \\ $$$$\Rightarrow \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{arctan}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}\:>\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{2}. \\ $$$$\mathrm{arctan}\:{u}\:+\mathrm{arctan}\:{v}\:=\mathrm{arctan}\:\frac{{u}+{v}}{\mathrm{1}−{uv}} \\ $$$$\:\:\:\:\:\left[\mathrm{for}\:\mathrm{some}\:\mathrm{values}\:\mathrm{of}\:{u},\:{v}\right] \\ $$$$ \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)\:=\frac{\mathrm{tan}\:\alpha\:+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta} \\ $$$$\Rightarrow \\ $$$$\mathrm{your}\:\mathrm{answers}\:\mathrm{are}\:\mathrm{wrong} \\ $$$$ \\ $$$$\mathrm{changing}\:\mathrm{the}\:\mathrm{question}\:\mathrm{to} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{arctan}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}} \\ $$$$\mathrm{makes}\:\mathrm{it}\:\mathrm{easilysolvable}\:\mathrm{because} \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}}=\frac{\left({n}+\mathrm{2}\right)−\left({n}+\mathrm{1}\right)}{\mathrm{1}+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$\Rightarrow \\ $$$$\mathrm{arctan}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}}\:=\mathrm{arctan}\:\left({n}+\mathrm{2}\right)\:−\mathrm{arctan}\:\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{but}\:\mathrm{with}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}\:\mathrm{this}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{work} \\ $$
Answered by Spillover last updated on 28/Dec/25
Rewritten using partial fractions (1/(n^2 +3n+2))=(1/(n+1))−(1/(n+2)) tan^(−1) ((1/(n+1))−(1/(n+2)))=tan^(−1) ((((n+2)−(n+1))/((1+(n+1)(n+2)))) tan^(−1) ((n+2)−(n+1))= S_N =Σ_(n=0) ^(N=0) tan^(−1) (n+2)−tan^(−1) (n+1) n=0,1,2,3... S_N =tan^(−1) (N+2)−tan^(−1) (1) S=lim_(N→∞) [tan^(−1) (N+2)−tan^(−1) (1)] S=(π/2)−(π/4)=(π/4)
$${Rewritten}\:{using}\:{partial}\:{fractions} \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}=\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\left({n}+\mathrm{2}\right)−\left({n}+\mathrm{1}\right)}{\left(\mathrm{1}+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\right.}\right) \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\left({n}+\mathrm{2}\right)−\left({n}+\mathrm{1}\right)\right)= \\ $$$${S}_{{N}} =\underset{{n}=\mathrm{0}} {\overset{{N}=\mathrm{0}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left({n}+\mathrm{2}\right)−\mathrm{tan}^{−\mathrm{1}} \left({n}+\mathrm{1}\right) \\ $$$${n}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}… \\ $$$${S}_{{N}} =\mathrm{tan}^{−\mathrm{1}} \left({N}+\mathrm{2}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$${S}=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{tan}^{−\mathrm{1}} \left({N}+\mathrm{2}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)\right] \\ $$$${S}=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$
Commented by Ghisom_ last updated on 29/Dec/25
Σ_(n=0) ^(10000) arctan (1/(n^2 +3n+2)) ≈.959>(π/4)
$$\underset{{n}=\mathrm{0}} {\overset{\mathrm{10000}} {\sum}}\mathrm{arctan}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}\:\approx.\mathrm{959}>\frac{\pi}{\mathrm{4}} \\ $$
Answered by mathmax last updated on 01/Jan/26
let n^2 +3n +2→ =9−8=1 ⇒ les racines n_1 =((−3+1)/2)=−1 et n_2 =((−3−1)/2)=−2 ⇒ n^2 +3n+2 =(n+1)(n+2) ⇒ Σ_(n=0) ^∞ arctan((1/(n^2 +3n+2)))=lim_(n→[) Σ_(k=0) ^n arctan(((k+2−(k+1))/(1+(k+1)(k+2)))) let n=tanu_n ⇒ Σ(...)=Σ_(k=0) ^n arctan(((tan(u_(k+2) )−tan(u_(k+1) ))/(1+tanu_(k+1) tan(u_(k+2) )))) =Σ_(k=0) ^n arctan(tan(u_(k+2) −u_(k+1) )) =Σ_(k=0) ^n (u_(k+2) −u_(k+1) ) =u_2 −u_1 +u_3 −u_2 +....+u_(n+2) −u_(n+1) =u_(n+2) −u_1 =arctan(n+1)−arctan(1) ⇒ limΣ(....)=(π/2)−(π/4)=(π/4)
$${let}\:{n}^{\mathrm{2}} +\mathrm{3}{n}\:+\mathrm{2}\rightarrow\:=\mathrm{9}−\mathrm{8}=\mathrm{1}\:\Rightarrow\:{les}\:{racines}\: \\ $$$${n}_{\mathrm{1}} =\frac{−\mathrm{3}+\mathrm{1}}{\mathrm{2}}=−\mathrm{1}\:{et}\:{n}_{\mathrm{2}} =\frac{−\mathrm{3}−\mathrm{1}}{\mathrm{2}}=−\mathrm{2}\:\Rightarrow \\ $$$${n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\:=\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{arctan}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}\right)={lim}_{{n}\rightarrow\left[\right.} \sum_{{k}=\mathrm{0}} ^{{n}} {arctan}\left(\frac{{k}+\mathrm{2}−\left({k}+\mathrm{1}\right)}{\mathrm{1}+\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}\right) \\ $$$${let}\:{n}={tanu}_{{n}} \:\Rightarrow \\ $$$$\Sigma\left(…\right)=\sum_{{k}=\mathrm{0}} ^{{n}} {arctan}\left(\frac{{tan}\left({u}_{{k}+\mathrm{2}} \right)−{tan}\left({u}_{{k}+\mathrm{1}} \right)}{\mathrm{1}+{tanu}_{{k}+\mathrm{1}} {tan}\left({u}_{{k}+\mathrm{2}} \right)}\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} {arctan}\left({tan}\left({u}_{{k}+\mathrm{2}} −{u}_{{k}+\mathrm{1}} \right)\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \left({u}_{{k}+\mathrm{2}} −{u}_{{k}+\mathrm{1}} \right) \\ $$$$={u}_{\mathrm{2}} −{u}_{\mathrm{1}} +{u}_{\mathrm{3}} −{u}_{\mathrm{2}} +….+{u}_{{n}+\mathrm{2}} −{u}_{{n}+\mathrm{1}} \\ $$$$={u}_{{n}+\mathrm{2}} −{u}_{\mathrm{1}} ={arctan}\left({n}+\mathrm{1}\right)−{arctan}\left(\mathrm{1}\right)\:\Rightarrow \\ $$$${lim}\Sigma\left(….\right)=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$

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