Using induction to prove a sequence is always less than a given number

Question

Let $f(1)=2$ and $f(n+1)=\sqrt{3+f(n)}$. Prove that $f(n)<2.4$ for all $n\ge 1$.

I established a base case when $n=1$ and then moved on to the inductive step by assuming the statement is true for some $k \in\mathbb N$. I am not sure what to do from here.

Answer 1

Start from $f(n) < 2.4$ $$3 + f(n) < 3+2.4$$ $$\sqrt{3+f(n)} < \sqrt{5.4}$$ $$f(n+1)<2.32 <2.4$$

Answer 2

1. $f(1)=2<2.4$.
2. $f(n)<2.4\implies f(n+1)=\sqrt{3+f(n)}<\sqrt{5.4}<2.4.$

So $f(1)<2.4\implies f(2)<2.4\implies f(3)<2.4\implies f(4)<2.4\implies f(5)<2.4\implies \cdots$

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It is interesting to note that if the sequence converges, it will converge to $f$ such that $f=\sqrt{3+f}$, or $f^2-f-3=0$.

$$f=(\sqrt{13}+1)/2\approx2.30277563773\cdots$$