Determine if the following sequence is monotonic

Question

Consider the recursion $$ X_1 = 10 \quad \&\quad X_{n+1} = \sqrt {3+2X_n}$$

Is it monotonic or not?

problem

What I have done so far.

$X_1 = 10$

$X_2 = \sqrt {23}$

$X_3 = \sqrt {2 \sqrt{23} +3}$

We can clearly see that the first term is bigger than the second and so on, so the sequence is actually decreasing. A great proof would be to prove that $(X_1-X_2)>0$ But you can't just do that by plugging in numbers.

I tried doing this so far.

I should get $(X_1-X_2)>0$ but I wind up getting $<0$. What's my mistake? work

Answer 1

One way to go about these type of questions, where the monotone property can't be seen directly by algebraic manipulation of the given recurrence relation, is using induction.

After computing some elements of the sequence we may have a guess if a sequence is monotonically increasing or decreasing. In our case, we see the elements get "smaller", so let's try proving it by induction.

Note that you already proved the base case: For $n=1$ we have $x_1=10$ and $x_2=\sqrt{23}$ and so $x_1>x_2$.

Now we establish our Induction Hypothesis (from now on IH), that is: We assume $x_{n-1}>x_n$ and prove that $x_n>x_{n+1}$. This is equivalent to $x_n-x_{n+1}>0$. So we compute (keeping in mind that we want to use our IH somehow):

$x_n=\sqrt{3+2x_{n-1}}$ and $x_{n+1}=\sqrt{3+2x_n}$, we now subtract to get:

$$x_n-x_{n+1}=\sqrt{3+2x_{n-1}}-\sqrt{3+2x_n}=\frac{2x_{n-1}-2x_n}{\sqrt{3+2x_{n-1}}+\sqrt{3+2x_n}}$$

(I've omitted some of the algebra since it's very similar to how you first approached this problem, which was very good!)

We now note the the denominator is always positive (as sum of two square roots), so the sign of $x_n-x_{n+1}$ will be determined by the numerator: $2x_{n-1}-2x_n$, but using our IH we know it's positive so $x_n-x_{n+1}>0$, and we proved by induction that the sequence is monotonically decreasing.

(I should mention that it's not always the most elegant or easy solution to these type of questions, but it's a solution that is easily generalized to other ambiguous cases, in which some trickery isn't always possible)

Answer 2

HINT: Note that $$x_{n+2}^2-x_{n+1}^2=2(x_{n+1}-x_n).$$ Now factor the LHS and since the sequence is clearly positive both $x_{n+2}-x_{n+1}$ and $x_{n+1}-x_n$ has the same parity for all $n\in \mathbb{N}.$ Then inductively $x_2-x_1$ must have the same parity, check it. For that comparing $x_2^2$ and $x_1^2$ is enough. (Why?)