# 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

##### Suggested Book

Mathematical Olympiad Challenges by Titu Andreescu & Razvan Gelca.

# 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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