There are 3 sequences $a_n $, $b_n$, $c_n$. Prove that limits in $\infty$ are equal and that 2 of them are convergent

Question

There are 3 sequences $a_n, $, $b_n$, $c_n$ with $n \in \mathbb{N} \setminus \{0\}$ given by recursions:

• $a_{n+1} = \frac{a_n+b_n+c_n}{3}$
• $b_{n+1} = \sqrt[3]{a_n*b_n*c_n}$
• $c_{n+1} = \frac{3}{\frac{1}{a_n} + \frac{1}{b_n}+\frac{1}{c_n}}$

Elements $a_1, b_1$ and $c_1$ are all positive numbers. Prove that at least $2$ of those $3$ sequences are convergent and that: $$\lim_{n \to +\infty} a_n = \lim_{n \to +\infty} b_n = \lim_{n \to +\infty} c_n $$

The first thing that I see is that those seqences are made of subsequent elements of functions known as arithmetic, geometric and harmonic means. Because of inequality of means, I know that: $$a_{n+1} = \frac{a_n+b_n+c_n}{3} \geq b_{n+1} = \sqrt[3]{a_n*b_n*c_n} \geq c_{n+1} = \frac{3}{\frac{1}{a_n} + \frac{1}{b_n}+\frac{1}{c_n}}$$

Then, I know for sure that:

• every subsequent element of $a_n$ must be smaller than the former one because $b_n$ and $c_n$ are smaller than $a_n$ for every n
• every subsequent element of $c_n$ must be greater than the former one because $a_n$ and $b_n$ are bigger than $c_n$ for every n

Additionally, I know that the limit of each one of those means is reached when all 3 elements in the mean are equal. In my case all 3 elements are the same for all 3 means so, the limit must be equal as well.

But here comes my problem - how to write it all down, to present my thinking in a correct way? Is there any scheme that I that could apply in my case (I only know how to formally find limits and prove convergence in cases of $\lim_{n \to +\infty}$ for sequences of $n$ elements)?

Answer 1

If you put $a_n \gt$ in front of your inequalities and $\gt c_n$ afterward you have a proof of convergence for $a$ and $c$ as the sequences are monotone and bounded.