Proof verification. Show $x_{n+1} = 1 + {b\over x_n}$ diverges if $b < -{1\over 4}$

Question

Given a recurrence relation: $$ x_{n+1} = 1 + {b\over x_n}\\ x_1 = a \in \Bbb R\\ n\in\Bbb N $$ Show that $x_n$ diverges if $b < -{1\over 4}$

I'm interested whether the below is a valid reasoning.

First suppose that $x_n$ converges. Then it must converge to some fixed points, which may be obtained by finding the roots of a characteristic polynomial.

So we are supposing: $$ \exists \lim_{n\to\infty}x_n = L $$ In case the limit exists then it must follow that: $$ L = 1 + {b\over L} \iff L^2-L-b=0 $$

The roots of that polynomial are given by: $$ L = \frac{1\pm \sqrt{1+4b}}{2} $$ Thus: $$ L \in \Bbb R \iff 1+4b>0 $$

So if $b < -{1\over 4}$ then $x_n$ is guaranteed to be divergent. However if $b \ge -{1\over 4}$ then the sequence may or may not converge.

Have I missed something?

Answer 1

I think you're right.

What you have done is that you searched for the necessary conditions in order to have the property "the sequence studied is converging". So if you don't have those conditions, it does not converge. And that is exactly the question.

Be cautious, (that wasn't the question) you didn't prove that it converge for the other cases. So the condition you find is sufficient but not necessary.

What you proved is that : $$ b < - \frac 1 4 \implies (x_n) \text{ diverges }$$

Answer 2

Yes, you proved divergence. There may be different modes of divergence -- blowing up without bound, or oscillating, or even having iterations appear to develop into a probability distribution over the real numbers (which is what you are really getting here, with most values of $a$ and $b<(-1/4)$). But anything other than convergence to a specific limit is duvergence, so proving the lack of such a specific limit here is all that is required.