Understand the problem

 For all natural numbers\(n\), let          \(A_n=\sqrt{2-\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}}\)           (\( n\) many radicals) (a) Show that for \(n\ge 2,  A_n=2\sin \frac{π}{2^{n+1}}\). (b) Hence, or otherwise, evaluate the limit                              \(\displaystyle\lim _{n\to \infty} 2^n A_n\) .

Source of the problem
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 6.
Topic
Trigonometric Substitution 

Difficulty Level
7.5 out of 10

Start with hints

Do you really need a hint? Try it first!

For \(n\ge 2\), \(\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}\)       (\( n\) many radicals) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2+0}}}\) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2+2\cos \frac{π}{2}}}}\).   

=\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2(1+\cos \frac{π}{2})}}}\) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{4\cos^2 \frac{π}{2^2}}}}\) =\(\sqrt{2+\sqrt{2+\cdots +2\cos \frac{π}{2^2}}}\). (\(n-1\) many radicals)      ……..    ………..     ……….      …   …….  …      …….     ………     …………    .….     ……… =\(2\cos \frac{π}{2^n}\)        \([n\ge 2]\).  

Now \(A_n=\sqrt{2-2\cos \frac{π}{2^n}}\) \(\Rightarrow  A_n= \sqrt{2(1-\cos \frac{π}{2^n})}\) \(\Rightarrow  A_n= \sqrt{4\sin^2 \frac{π}{2^{n+1}}}\)    \(\Rightarrow  A_n= 2\sin \frac{π}{2^{n+1}}\). 

Now \(\displaystyle\lim_{n\to \infty} 2^nA_n=\displaystyle\lim_{n\to \infty} 2^{n+1}\cdot \sin \frac{π}{2^{n+1}}\). Since , \(n \to \infty \) \(\Rightarrow 2^{n+1}\to \infty \) \(\Rightarrow \frac{π}{2^{n+1}}\to 0\) Let \(\frac{π}{2^{n+1}}=z \Rightarrow z \to 0\), when \(n \to \infty\). Therefore, \(\displaystyle\lim_{n\to \infty} 2^n A_n \)=\(\displaystyle\lim_{z \to 0}\frac{\sin z}{z}\cdot π =1\times π=π\). (Ans.)               

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