Recurring Styles #14 - Nested Radicals!

Calculus Level 5

$\large{ x_{n+1} = \sqrt{2+x_n} \quad, \quad x_1 = \sqrt{2} }$

Consider the sequence $(x_n)_{n \in \mathbb Z^+}$ as defined above. Let $L = \displaystyle \lim_{n \to \infty} 4^n(2-x_n)$ . If $L$ can be expressed in the form $\dfrac{A}{B} \pi^C$ for positive integers $A,B,C$ , find the minimum value of $A+B+C$ .

Bonus: Generalize $x_n$ in terms of $n$ .

The answer is 7.

1 solution

Otto Bretscher
Sep 11, 2015

By the half-angle formula for the cosine we have $x_n=2\cos\left(\frac{\pi}{2^{n+1}}\right)$ . Now $\lim_{n\to\infty}4^n(2-x_n)$ $=\lim_{n\to\infty}4^n\left(2-2\cos\left(\frac{\pi}{2^{n+1}}\right)\right)=\lim_{n\to\infty}4^{n+1}\sin^2\left(\frac{\pi}{2^{n+2}}\right)=\lim_{n\to\infty}4^{n+1}\frac{\pi^2}{4^{n+2}}=\frac{\pi^2}{4}$

Moderator note:

Nice observation of the half-angle trig substitution.

sir, did something like this: $x_n=\sqrt{2+x_{n-1}}=\sqrt{2+\sqrt{2+x_{n-2}}=\sqrt{2+\sqrt{2+\sqrt{2+...}}}(n-1 times)$ so, $lim_(n\rightarrow\infty)(x_n)=\sqrt{2+\sqrt{2+\sqrt{2+...}}}=2$ and $lim_{n\rightarrow \infty} (4^n(2-2))=0$ what is wrong?

Aareyan Manzoor - 5 years, 8 months ago

1 thing approaches infinity, the other approaches 0. The limit does not necessarily approach 0.

Joe Mansley - 4 years, 6 months ago

Recurring Styles #14 - Nested Radicals!

The answer is 7.

1 solution

Moderator note:

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