 ## 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

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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