# Determine if the following sequence is monotonic

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

• Welcome to Mathematics Stack Exchange. Please use Mathjax. $x>3\implies(x-3)(x+1)>0\implies\sqrt{3+2x}<x$ – J. W. Tanner Sep 16 '19 at 22:51
• The linked problem only asks if the recursion is monotonic. It doesn't specify increasing or decreasing. – lulu Sep 16 '19 at 22:51
• @lulu Yes, but can I know if a sequence is monotonic, without knowing if it is decreasing/increasing/bounded/unbounded? – Arman Babayan Sep 16 '19 at 22:53
• I don't understand. As you have remarked, $X_2<X_1$. Therefore the sequence starts out as decreasing. To prove that it is monotonic therefore requires you to prove that it is decreasing for all $n$. That is, prove that $X_{n+1}<X_n$ for all $n$ (or find a counterexample). – lulu Sep 16 '19 at 22:55
• Yes you are correct, but it requires a real proof. I cannot just say that 10 is bigger than sqrt 23 and be done with it, because maybe this does not work for all the values of the sequence. – Arman Babayan Sep 16 '19 at 22:57

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)

• Thank you so much, best explanation. – Arman Babayan Sep 17 '19 at 7:26

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?)