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# 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

7.5 out of 10

##### Suggested Book

Mathematical Olympiad Challenges by Titu Andreescu & Razvan Gelca.

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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