Six six six

Calculus Level 2

Evaluate the infinitely nested radical expression, $\sqrt{6 + \sqrt{6 + \sqrt{ 6 + \sqrt{ 6 + \cdots}}}} .$

The answer is 3.

2 solutions

Ngọc Bùi Tiến
Mar 10, 2015

$x=\sqrt{6+x}$

$x^2=6+x$

$x^2-x-6=0$

$(x-3)(x+2)=0 \rightarrow x=3$ or $x=-2$

Since $x>0$ , the only answer is $\boxed{x=3}$

You can also view tricky problems of this kind in my set Nested Radicals.....

A Former Brilliant Member - 6 years, 3 months ago

I must ask that old question: But how do we know that it converges in the first place? The solution is not complete, and the most interesting and important part is missing.

Otto Bretscher - 2 years, 6 months ago

Let we have a series: $f(1) = \sqrt{6}$ , $f(n) = \sqrt{6 + f(n - 1)}$ . It's clear that the series is increasing (the (n-1)th square root of f(n) is $6+\sqrt{6}$ while the (n-1)th square root of f(n-1) is 6) and that the series has a limit says, 30 (because f(1) < 30 and f(n+1) < 30 if f(n) <30). An increasing series with a limit is converging.

Ngọc Bùi Tiến - 2 years, 6 months ago

Totally easy question :)

Frankie Fook - 6 years, 3 months ago

Otto Bretscher
Nov 16, 2018

Define the function $f(x)=\sqrt{6+x}$ , for $x\geq -6$ , and consider the sequence $a_n$ recursively defined by $a_0=0$ and $a_{n+1}=f(a_n)$ . The value of the nested radical is $\lim_{n \to \infty}a_n$ , if indeed it exists.

We observe that $f(x)$ is increasing, $f(3)=3$ and $f(x)>x$ for $x<3$ . It follows that $a_n$ is an increasing sequence, bounded by 3, that must converge to the only fixed point of $f(x)$ , namely, $x=\boxed{3}$ .

Six six six

The answer is 3.

2 solutions

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