Find the value of $\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}$

Question

Evaluate

$$\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}$$

I need some help with this question because I have no idea what is going on and help would be greatly appreciate :)

Answer 1

Put $$ x=\sqrt{6+\sqrt{6+\sqrt{6+\dots}}} $$ squaring you get $$ x^2=6+\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}=6+x\;\;\;. $$ Thus you simply have to solve $$ x^2-x-6=0\;\; $$ which has two solutions: $x=-2$ and $x=3$. But $x$ is clearly positive, thus you can conclude that $$ \sqrt{6+\sqrt{6+\sqrt{6+\dots}}}=3\;. $$

Answer 2

$\sqrt{6+\sqrt{6+\sqrt{6+\dots}}} = \lim x_n$ where $x_{n+1} = \sqrt{6+x_n}$ and $x_0 \ge 0$.

As others have explained, if this sequence converges, then it converges to $3$.

To prove that it does converge, prove this:

• If $x_0 < 3 $ then the sequence is increasing and bounded above by $3$.

• If $x_0 > 3 $ then the sequence is decreasing and bounded below by $3$.

In both cases, you get a monotone bounded sequence, and so it converges.