Find-lim-n-n-pi-2-4-a-n-2-where-a-n-k-1-n-arctan-1-k-2-k-1-

Question Number 153571 by mathdanisur last updated on 08/Sep/21

Find lim_(n→∞) n∙((π^2 /4) - a_n ^2 ) = ? where a_n =Σ_(k=1) ^n arctan((1/(k^2 -k+1)))

$$\mathrm{Find}\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}n}\centerdot\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:-\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} \right)\:=\:? \\ $$$$\mathrm{where}\:\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} =\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} -\mathrm{k}+\mathrm{1}}\right) \\ $$

Answered by mnjuly1970 last updated on 08/Sep/21

solution.. a_( n) = Σ_(k=1 ) ^n arctan(((k −(k−1))/(1+k (k−1)))) =Σ_(k=1) ^n {arctan(k) − arctan(k−1)} = arctan(n ) ∴ lim_( n→∞) n.((π^2 /4) −arctan^2 (n)) = lim_( n→∞) n.((π/2) −arctan(n))((π/2) +arctan(n)) = lim_(n→∞) n.arccot(n).lim_(n→∞) ((π/2) +arctan(n)) =πlim_( n→∞) n.arccot(n)= π.1=π

$$\:\:{solution}.. \\ $$$$\:\:\:\:\:{a}_{\:{n}} =\:\underset{{k}=\mathrm{1}\:} {\overset{{n}} {\sum}}{arctan}\left(\frac{{k}\:−\left({k}−\mathrm{1}\right)}{\mathrm{1}+{k}\:\left({k}−\mathrm{1}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left\{{arctan}\left({k}\right)\:−\:{arctan}\left({k}−\mathrm{1}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{arctan}\left({n}\:\right) \\ $$$$\:\:\:\:\:\:\therefore\:{lim}_{\:{n}\rightarrow\infty} \:{n}.\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:−{arctan}^{\mathrm{2}} \left({n}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:{lim}_{\:{n}\rightarrow\infty} {n}.\left(\frac{\pi}{\mathrm{2}}\:−{arctan}\left({n}\right)\right)\left(\frac{\pi}{\mathrm{2}}\:+{arctan}\left({n}\right)\right) \\ $$$$\:\:=\:{lim}_{{n}\rightarrow\infty} \:{n}.{arccot}\left({n}\right).{lim}_{{n}\rightarrow\infty} \left(\frac{\pi}{\mathrm{2}}\:+{arctan}\left({n}\right)\right) \\ $$$$\:\:\:\:=\pi{lim}_{\:{n}\rightarrow\infty} {n}.{arccot}\left({n}\right)=\:\pi.\mathrm{1}=\pi \\ $$$$ \\ $$

Commented by mathdanisur last updated on 08/Sep/21