Is k=$\sqrt{6+\sqrt{6+...}}$ monotonically increasing?

Question

Is k=$\sqrt{6+\sqrt{6+...}}$ monotonically increasing or is not ? i know it converges to $3$.

Answer 1

We can induct on the amount of nested roots.

Base Case:

$\sqrt{6} < \sqrt{6+\sqrt{6}} \implies 6 < 6 + \sqrt6 \implies 0 < \sqrt6$

Inductive step: Suppose that $\sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n-1} \ times}} < \sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n} \ times}}$. We prove: $\sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n} \ times}} < \sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n+1} \ times}}$.

$\sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n} \ times}} < \sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n+1} \ times}} \implies 6+\underbrace{\sqrt{6+...}}_{\text{n} \ times} < 6+\underbrace{\sqrt{6+...}}_{\text{n+1} \ times} \implies \underbrace{\sqrt{6+...}}_{\text{n} \ times} < \underbrace{\sqrt{6+...}}_{\text{n+1} \ times} \implies \sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n-1} \ times}} < \sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n} \ times}}$

So, by the inductive hypothesis, we have that $k_n < k_{n+1}$ for all $n$. Thus, the sequence $k$ is strictly monotonically increasing.

Please tell me if there is anything wrong with this solution.

EDIT

I proved this in reverse. The steps are equally valid in reverse, so the proof is essentially right anyway. In any case, here is an adapted version with a little explanation.

Suppose we have: $\sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n-1} \ times}} < \sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n} \ times}}$. This can just as well be written as: $\underbrace{\sqrt{6+...}}_{\text{n} \ times} < \underbrace{\sqrt{6+...}}_{\text{n+1} \ times}$. Adding $6$ to both sides changes nothing, so we get: $6+\underbrace{\sqrt{6+...}}_{\text{n} \ times} < 6+\underbrace{\sqrt{6+...}}_{\text{n+1} \ times}$. We can then take the square root of both sides without worrying about flipping the inequality since both sides are clearly positive, this yields: $\sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n} \ times}} < \sqrt{6+\underbrace{\sqrt{6+...}}_{\text{n+1} \ times}}$, as desired.

Thus, by the inductive hypothesis, $k$ is strictly monotonically increasing.