real analysis - Prove the sequence $x_{n+1}=frac{1}{4-x_n}, x_1=3$ converges.

I try to use induction to show that the sequence is monotonically decreasing. My induction hypothesis is $$P(n):x_{n+1}
We have first few terms $x_1=3, x_2=1, x_3=1/3,\cdots$. It is clear that $P(1)$ and $P(2)$ hold. Now I try to prove $P(n+1)$ assuming $P(n)$ is true.
To do this I must show that $x_{n+2}-x_{n+1}<0$.

$$x_{n+2}-x_{n+1}=\frac{1}{4-x_{n+1}}-\frac{1}{4-x_n}=\frac{x_{n+1}-x_n}{(4-x_{n+1})(4-x_n)}$$
I get stuck here I know that numerator is less than $0$ but how do I show that denominator is positive so that the entire thing is less than $0$.

I am trying to show that sequence is monotonically decresaing and also bounded below to show that it converges.

Hints first

Answer

Hint: use as your inductive hypothesis the following statement: $x_{n+1} < x_n < 4$.

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How to come up with this hypothesis? I first asked myself whether you were asking for something that's even true: is the denominator positive? The answer is obviously yes by working out the first few values. Why is the denominator positive? Because it's the product of positive things. Can we show that we're always taking the product of positive things? In order to do that, we'd need…