Question Number 127501 by slahadjb last updated on 30/Dec/20
$${prove}\:{the}\:{convergence}\:{of}\:\left(\alpha_{{n}} \right)_{{n}\:} {such}\:{that}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\alpha_{{n}} }{{n}}\right)−\mathrm{2}\alpha_{{n}} +\mathrm{1}=\mathrm{0} \\ $$
Answered by mindispower last updated on 31/Dec/20
![⇒2a_n =1+tan^− ((a_n /n))⇒a_n ∈[−(π/4)+(1/2),(π/4)+(1/2)] bounded⇒lim_(n→∞) (a_n /n)→0 ⇒tan^− ((α_n /n))+1⇒1sinceg: x→tan^(−1) (x)+1 is continus 2a_n =g((a_n /n)),2a_n →g(0)=1 a_n →(1/2)](Q127599.png)
$$\Rightarrow\mathrm{2}{a}_{{n}} =\mathrm{1}+{tan}^{−} \left(\frac{{a}_{{n}} }{{n}}\right)\Rightarrow{a}_{{n}} \in\left[−\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}},\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$${bounded}\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{a}_{{n}} }{{n}}\rightarrow\mathrm{0} \\ $$$$\Rightarrow{tan}^{−} \left(\frac{\alpha_{{n}} }{{n}}\right)+\mathrm{1}\Rightarrow\mathrm{1}{sinceg}:\:{x}\rightarrow\mathrm{tan}^{−\mathrm{1}} \left({x}\right)+\mathrm{1}\:{is}\:{continus} \\ $$$$\mathrm{2}{a}_{{n}} ={g}\left(\frac{{a}_{{n}} }{{n}}\right),\mathrm{2}{a}_{{n}} \rightarrow{g}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${a}_{{n}} \rightarrow\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by slahadjb last updated on 31/Dec/20
$${Thank}\:{you}.\:{Can}\:{we}\:{prove}\:{that}\:\left(\alpha_{{n}} \right)\:{is}\:{monotone}\:? \\ $$
Commented by mindispower last updated on 31/Dec/20
$$\mathrm{2}{a}_{{n}} =\mathrm{1}+{tan}^{−} \left(\frac{{a}_{{n}} }{{n}}\right) \\ $$$$\mathrm{1}=\mathrm{2}{a}_{{n}} −{tan}^{−} \left(\frac{{an}}{{n}}\right) \\ $$$${f}\left({x}\right)=\mathrm{2}{x}−{tan}^{−} \left(\frac{{x}}{{n}}\right), \\ $$$$\mathrm{2}{a}_{{n}} −{tan}^{−} \left(\frac{{a}_{{n}} }{{n}+\mathrm{1}}\right)=\mathrm{1}+{tan}^{−} \left(\frac{{a}_{{n}} }{{n}}\right)−{tan}^{−} \left(\frac{{a}_{{n}} }{{n}+\mathrm{1}}\right)>\mathrm{1} \\ $$$${a}_{{n}} >\mathrm{0},\forall{n}\geqslant\mathrm{4} \\ $$$$\geqslant\mathrm{2}{a}_{{n}+\mathrm{1}} −{tan}^{−} \left(\frac{{a}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\right)={f}\left({a}_{{n}+\mathrm{1}} \right) \\ $$$${f}\left({a}_{{n}} \right)\geqslant{f}\left({a}_{{n}+\mathrm{1}} \right)\Rightarrow{a}_{{n}} >{a}_{{n}+\mathrm{1}} \:\:{n}>\mathrm{4},{causef}\:\:{is}\:{increase} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by slahadjb last updated on 31/Dec/20
$${thank}\:{you}\:{so}\:{much}. \\ $$
Commented by mindispower last updated on 31/Dec/20
$${pleasur} \\ $$