Great question! Let's work through this step by step to show all three sequences converge to the same limit. We'll rely on the Monotone Convergence Theorem (for bounded, monotonic sequences) and algebraic manipulation to connect the pieces.

Step 1: Prove $(a_n)$ and $(c_n)$ are Convergent #

First, let's analyze the arithmetic mean sequence $(a_n)$:

• For every $n$, since $a_n \le b_n \le c_n$, the arithmetic mean $a_{n+1} = \frac{a_n + b_n + c_n}{3}$ satisfies $a_n \le a_{n+1} \le c_n$. This means $(a_n)$ is monotonically increasing (each term is at least the previous one).
• Also, we'll soon see $(c_n)$ is decreasing, so $a_n \le c_1$ for all $n$—giving $(a_n)$ an upper bound. By the Monotone Convergence Theorem, $(a_n)$ converges to some limit $\alpha$.

Next, the harmonic mean sequence $(c_n)$:

• The harmonic mean $c_{n+1} = \frac{3}{\frac{1}{a_n} + \frac{1}{b_n} + \frac{1}{c_n}}$. Since $a_n \le b_n \le c_n$, their reciprocals satisfy $\frac{1}{a_n} \ge \frac{1}{b_n} \ge \frac{1}{c_n}$. The arithmetic mean of these reciprocals is $\frac{\frac{1}{a_n} + \frac{1}{b_n} + \frac{1}{c_n}}{3} \ge \frac{3}{3c_n} = \frac{1}{c_n}$; taking reciprocals reverses the inequality, so $c_{n+1} \le c_n$. Thus $(c_n)$ is monotonically decreasing.
• Since $(a_n)$ is increasing, $c_n \ge a_n \ge a_1$ for all $n$—giving $(c_n)$ a lower bound. Again by the Monotone Convergence Theorem, $(c_n)$ converges to some limit $\gamma$.

Step 2: Convergence of $(b_n)$ via the Squeeze Theorem #

For the geometric mean sequence $(b_n)$:

• By induction, $a_n \le b_n \le c_n$ holds for all $n$: the base case $n=1$ is given. Assume $a_n \le b_n \le c_n$; the mean inequality tells us $A(a_n,b_n,c_n) \ge B(a_n,b_n,c_n) \ge C(a_n,b_n,c_n)$, so $a_{n+1} \ge b_{n+1} \ge c_{n+1}$. Additionally, $b_{n+1} = (a_n b_n c_n)^{1/3} \ge (a_n³⁾{1/3} = a_n$ and $b_{n+1} \le (c_n³⁾{1/3} = c_n$, so $a_n \le b_n \le c_n$ holds for all $n$.
• Since $a_n \to \alpha$ and $c_n \to \gamma$, if we can show $\alpha = \gamma$, the squeeze theorem will force $b_n \to \alpha$ as well.

Step 3: Show $\alpha = \gamma$ (All Limits Are Equal) #

Let $\beta$ denote the limit of $(b_n)$ (we know it exists if $\alpha = \gamma$, but let's use the recurrence relations directly):

1. Take the limit of the arithmetic mean recurrence:
   $$\alpha = \frac{\alpha + \beta + \gamma}{3}$$
   Multiply through by 3:
   $$3\alpha = \alpha + \beta + \gamma \implies 2\alpha = \beta + \gamma \tag{1}$$

2. Take the limit of the harmonic mean recurrence:
   $$\gamma = \frac{3}{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}}$$
   Multiply numerator and denominator by $\alpha\beta\gamma$ to simplify:
   $$\gamma = \frac{3\alpha\beta\gamma}{\alpha\beta + \beta\gamma + \gamma\alpha}$$
   Since all terms are positive, divide both sides by $\gamma$:
   $$1 = \frac{3\alpha\beta}{\alpha\beta + \beta\gamma + \gamma\alpha}$$
   Cross-multiply to get:
   $$\alpha\beta + \beta\gamma + \gamma\alpha = 3\alpha\beta \implies \beta\gamma + \gamma\alpha = 2\alpha\beta \tag{2}$$

3. Substitute $\beta = 2\alpha - \gamma$ (from equation 1) into equation 2:
   $$(2\alpha - \gamma)\gamma + \gamma\alpha = 2\alpha(2\alpha - \gamma)$$
   Expand and simplify:
   $$2\alpha\gamma - \gamma^2 + \alpha\gamma = 4\alpha^2 - 2\alpha\gamma$$
   $$3\alpha\gamma - \gamma^2 = 4\alpha^2 - 2\alpha\gamma$$
   Rearrange into a quadratic equation:
   $$\gamma^2 -5\alpha\gamma +4\alpha^2=0$$
   Factor this:
   $$(\gamma - \alpha)(\gamma -4\alpha)=0$$

The solutions are $\gamma = \alpha$ or $\gamma = 4\alpha$. The second solution is impossible: substituting $\gamma=4\alpha$ into equation (1) gives $\beta = 2\alpha -4\alpha = -2\alpha$, but $\beta$ is the limit of positive terms $b_n$, so it can't be negative. Thus $\gamma = \alpha$, and substituting back into equation (1) gives $\beta = \alpha$.

Conclusion #

• $(a_n)$ converges to $\alpha$, $(c_n)$ converges to $\alpha$, and $(b_n)$ is squeezed between them so it also converges to $\alpha$. All three sequences therefore converge to the same limit.

内容的提问来源于stack exchange，提问作者MOP