Question Number 179917 by mnjuly1970 last updated on 04/Nov/22
Answered by a.lgnaoui last updated on 06/Nov/22
![(1/(4^n cos^2 ((π/2^(n+2) ))))=(1/4^n )(1+tan^2 ((π/2^(n+2) )) tan (x)≅x+(x^3 /3)+(2/(15))x^5 x^2 (1+(x^2 /3))^2 <tan^2 (x)<x^2 (1+(x^2 /3)+(2/(15))x^4 )^2 x^2 +(2/3)x^4 +(x^6 /9)<tan^2 (x)<x^2 +((2x^4 )/3)+(x^6 /9)+((2x^6 )/(15)) (1/4^n )[(1+((π/2^(n+2) ))^2 +(2/3)((π/2^(n+2) ))^4 +((π/2^(n+2) ))^6 ] <(1/4^n )[1+tan^2 ((π/2^(n+2) ))]<(1/4^n )[1+((π/2^(n+2) ))^2 +(2/3)((π/2^(n+2) ))^4 +((11)/(45))(^6 (π/2^(n+2) ))^6 ] (1/2^(2n) )+(π^2 /2^(4n+4) )+((2^4 π^4 )/(3×2^(6n+8) ))<S<(1/2^(2n) )+(π^2 /2^(4n+4) )+(2/3)×(π^4 /2^(6n+8) )+(((11)/(45))×(π^6 /2^(8n+12) )) ⇒lim_(n→∞) Σ_(n=1) ^∞ (1/4^n )×(1/(cos^2 ((π/2^(n+2) ))))=lim_(n→∞) (1/2^(2n) )=0](Q180019.png)
$$\:\:\:\frac{\mathrm{1}}{\mathrm{4}^{{n}} \mathrm{cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{2}} }\right)}=\frac{\mathrm{1}}{\mathrm{4}^{{n}} }\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{2}} }\right)\right. \\ $$$$\mathrm{tan}\:\left({x}\right)\cong{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{15}}{x}^{\mathrm{5}} \:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)^{\mathrm{2}} <\mathrm{tan}^{\mathrm{2}} \left({x}\right)<{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{15}}{x}^{\mathrm{4}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{4}} +\frac{{x}^{\mathrm{6}} }{\mathrm{9}}<\mathrm{tan}^{\mathrm{2}} \left({x}\right)<{x}^{\mathrm{2}} +\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{3}}+\frac{{x}^{\mathrm{6}} }{\mathrm{9}}+\frac{\mathrm{2}{x}^{\mathrm{6}} }{\mathrm{15}} \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{4}^{{n}} }\left[\left(\mathrm{1}+\left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{2}} }\right)^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{2}} }\right)^{\mathrm{4}} \:+\left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{2}} }\right)^{\mathrm{6}} \right]\:\:<\frac{\mathrm{1}}{\mathrm{4}^{{n}} }\left[\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{2}} }\right)\right]<\frac{\mathrm{1}}{\mathrm{4}^{{n}} }\left[\mathrm{1}+\left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{2}} }\right)^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{2}} }\right)^{\mathrm{4}} +\frac{\mathrm{11}}{\mathrm{45}}\left(^{\mathrm{6}} \frac{\pi}{\mathrm{2}^{{n}+\mathrm{2}} }\right)^{\mathrm{6}} \right]\right. \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }+\frac{\pi^{\mathrm{2}} }{\mathrm{2}^{\mathrm{4}{n}+\mathrm{4}} }+\frac{\mathrm{2}^{\mathrm{4}} \pi^{\mathrm{4}} }{\mathrm{3}×\mathrm{2}^{\mathrm{6}{n}+\mathrm{8}} }<{S}<\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }+\frac{\pi^{\mathrm{2}} }{\mathrm{2}^{\mathrm{4}{n}+\mathrm{4}} }+\frac{\mathrm{2}}{\mathrm{3}}×\frac{\pi^{\mathrm{4}} }{\mathrm{2}^{\mathrm{6}{n}+\mathrm{8}} }+\left(\frac{\mathrm{11}}{\mathrm{45}}×\frac{\pi^{\mathrm{6}} }{\mathrm{2}^{\mathrm{8}{n}+\mathrm{12}} }\right) \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{4}^{{n}} }×\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{2}} }\right)}=\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }=\mathrm{0} \\ $$