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 $a_3 \notin \Bbb{R}$?

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

2 Answers 2


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

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

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


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