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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

Ser, = ∞  or  π.?

$$\boldsymbol{\mathrm{S}}\mathrm{er},\:=\:\infty\:\:\mathrm{or}\:\:\pi.? \\ $$

Commented by mathdanisur last updated on 08/Sep/21

Ser,  arctan(n)  or  arccot(n).?

$$\boldsymbol{\mathrm{S}}\mathrm{er},\:\:\mathrm{arctan}\left(\mathrm{n}\right)\:\:\mathrm{or}\:\:\mathrm{arccot}\left(\mathrm{n}\right).? \\ $$

Commented by mnjuly1970 last updated on 08/Sep/21

     arccotan(n)

$$\:\:\:\:\:{arccotan}\left({n}\right) \\ $$

Commented by mathdanisur last updated on 08/Sep/21

Very nice thankyou ser

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{thankyou}\:\mathrm{ser} \\ $$

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