Nested Radicals

Calculus Level 2

$\sqrt{6}, \sqrt{6+\sqrt{6}}, \sqrt{6+\sqrt{6 + \sqrt{6}}}, \ldots$

What does this sequence converge to?

The answer is 3.

1 solution

Mateus Gomes
Feb 5, 2016

$\ {\sqrt{6+\color{#3D99F6}{\sqrt{6+\sqrt{\cdots+\sqrt{6}}}}}}= \, {\color{#3D99F6}{x}}$ Squring both sides. $\color{#3D99F6}{x}^2=6+\color{#3D99F6}{x}$ $\color{#3D99F6}{x}^2-\color{#3D99F6}{x}-6=0$ $\color{#3D99F6}{x}=-2$ $\color{#3D99F6}{x}=\color{#D61F06}{\boxed 3}$ (Note that $x\geqslant {0}$ .)

You need to show that the sequence converges to 3.

Otto Bretscher - 5 years, 4 months ago

Proof by induction that $a_n$ is bounded above by $3$ : $a_1=sqrt(6) < 3$ . If $a_n \leq 3$ , then $a_{n+1} = sqrt(a_n+6) \leq sqrt(6 + 3) = 3$ . Proof that $a_n$ is monotonically increasing: If $a_n \leq 3$ , then $a_n^2 \leq 6 + a_n \leq a_{n+1}^2$ . Therefor $a_{n+1} \geq a_n$ .

Maximilian Wackenhuth - 5 years, 4 months ago

You got the right idea, but your proof needs some work. For example, why is $a_n$ "trivially" increasing"? If $a_n=4$ , then $a_{n+1}=\sqrt{10}<4$ . You want to show that $a_n$ is bounded above by 3, not just by 6.

Otto Bretscher - 5 years, 4 months ago

Nested Radicals

The answer is 3.

1 solution

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