# Is the sequence defined by $a_{1}=\frac43, a_{n+1} = \sqrt{5a_{n}-6}$ monotonically decreasing?

I want to know where is my mistake in the proof by induction that $$(a_n)$$ is monotonically decreasing.

Basis of induction: $$a_{2} - a_{1} = \sqrt{\dfrac{2}{3}} - \dfrac43 < 0$$

Assume $$a_{n+1} - a_{n} < 0$$

Then $$a_{n+2} - a_{n+1} = \sqrt{5a_{n+1}-6} - \sqrt{5a_{n}-6}= ( \sqrt{5a_{n+1}-6}- \sqrt{5a_{n}-6})\dfrac{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}=\dfrac{5a_{n+1}-6-5a_n+6}{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}=\dfrac{5(a_{n+1}-a_n)}{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}<0$$

So $$a_{n+1} - a_n < 0 \;\;\forall n \in \Bbb{N}$$, hence this sequence is monotonically decreasing.

But how can this be true if $$a_3 \notin \Bbb{R}$$?

• Don't you mean after $n=8$ the real part is decreasing and after $n=4$ the imaginary part is decreasing?
– user561334
Dec 18, 2020 at 11:16
• I meant only the latter part, I forgot about the real part. I have deleted the previous comment. Dec 18, 2020 at 12:53

A sequence is a function from the natural numbers to the real line, that is you have $$a_n: \mathbb{N} \rightarrow \mathbb{R}$$, where the pedex $$n$$ is used as $$x$$ in the standard $$f(x)$$ notation. This sequence is not well defined if $$a_n < 6/5$$, but for the values with $$a_n \ge 6/5$$ you can apply your proof, that basically means your sequence is not well defined for every $$n \ge 3$$, since you proved inductively that increasing $$n$$, for sure you do not get a higher value for $$a_n$$. Since the definition of monotonically decreasing sequence is that $$a_{n+1} \le a_n$$ for all $$n \in \mathbb{N}$$, your sequence is not monotonically decreasing.

With appropriate initial values, you can have $$a_n \in \mathbb R, \forall n \in \mathbb N$$. In this case you can do better than writing all those ugly squareroots.

You can prove easily that $$a_n \ge 2 \implies a_n \in \mathbb R, \forall n \in \mathbb N$$ and

• If $$a_1=2$$ or $$3$$ then $$a_n \equiv a_1$$.
• If $$2 < a_1 < 3$$, then $$2 < a_n < 3$$.
• If $$a_1 > 3$$, then $$a_n > 3$$.

The trick is to square your equation and subtract the square of one of the stationary points, $$2$$ or $$3$$. Let's do $$2$$.

$$a_{n+1}^2-4 = 5a_n-6-4 \implies (a_{n+1}+2)(a_{n+1}-2)=5(a_n-2)$$

Assume $$a_1 > 2$$ (thus $$a_n>2, \forall n \in \mathbb N$$). Then

$$a_1 < 3 \implies a_n < 3 \\ \implies 5(a_n-2) = (a_{n+1}+2)(a_{n+1}-2)< 5(a_{n+1}-2)\\ \implies a_n < a_{n+1}$$

$$a_1 > 3 \implies a_n < 3 \\ \implies 5(a_n-2) = (a_{n+1}+2)(a_{n+1}-2) > 5(a_{n+1}-2)\\ \implies a_n > a_{n+1}$$