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

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

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

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: $$a_{n+1} = \frac{a_n+b_n+c_n}{3} \geq b_{n+1} = \sqrt{a_n*b_n*c_n} \geq c_{n+1} = \frac{3}{\frac{1}{a_n} + \frac{1}{b_n}+\frac{1}{c_n}}$$

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)?

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.
• You said that $a_{n+1} \lt a_n$, which says the $a$ sequence is monotone decreasing and that $c$ is monotone increasing. Nov 22, 2020 at 16:51