proving f(n) is strictly less than 2.4 when $n\geq 1$

Question

$f(1)= 2, f(n+1) = \sqrt{3+f(n)}$.

Prove $f(n) < 2.4$ for all $n ≥ 1$.

Would this be a proof by induction? If so, could somebody start me off?

Answer 1

Let's do these passages to prove this.

1) $f(n) \geq 0$

2) $f(n+1) \geq f(n)$

3) $f(n) \rightarrow l \in \overline{\mathbb{R}}$

4) $f(n) \rightarrow \frac{1+\sqrt{13}}{2}$

You prove the first two passages by induction, then you conclude the third with the monotone convergence theorem.

For the last point, which is the correct limit of your succession, let's think this: compute the fixed point of this function, then it must go there.

So you just do $x=f(x)$ and find the solution of this equations. These are the possible limits and you choose the one you are interested into by seeing how this function goes.

Answer 2

You already have the "first step", f(1)= 2< 2.4.

Now, suppose that, for some k, f(k)< 2.4. Then $f(k+1)= \sqrt{3+ f(k)}< \sqrt{3+ 2.4}= \sqrt{5.4}$. How large is $\sqrt{5.4}$?

Answer 3

Hint: $\alpha=\frac{1}{2}\left(1+\sqrt{13}\right)$ is the only solution of $x=\sqrt{3+x}$. If you prove that

$$ \forall x\geq 2,\qquad \left|\sqrt{3+x}-\alpha\right|\leq \left|x-\alpha\right| \tag{1} $$ it follows that for every $n\geq 3$ $$ f(n)\in [2,\alpha]\tag{2} $$ since $\sqrt{3+x}\geq x$ on $[2,\alpha]$. You may notice that $(1)$ can be written as $$\left|\frac{\sqrt{3+x}-\sqrt{3+\alpha}}{x-\alpha}\right|\leq 1 \tag{3} $$ for any $x\neq\alpha$, and that $(1)$ follows from the Lipshitz-continuity of $\sqrt{3+x}$ over $[2,+\infty)$.
Actually $(1)$ can be improved by much: $$\forall x\geq 2,\qquad \left|\sqrt{3+x}-\alpha\right|\leq\color{red}{\frac{1}{2\sqrt{5}}}\left|x-\alpha\right|.\tag{1improved}$$