证明递归定义序列有界且不收敛，验证子序列极限

Alright, let's work through this problem step by step. We've got a sequence defined by $x_1 = \frac{2}{3}$, with recursive rules:

• $x_{2n+1} = \frac{x_{2n}}{3} + \frac{2}{3}$
• $x_{2n} = \frac{x_{2n-1}}{3}$

You've already split it into odd and even subsequences, which is a great start—let's formalize the proofs for boundedness and non-convergence.

1. Prove the sequence is bounded ($0 < x_n < 1$ for all $n$) #

We can use mathematical induction here to cover all terms:

• Base case: For $n=1$, $x_1 = \frac{2}{3}$, which clearly sits between 0 and 1. For $n=2$, $x_2 = \frac{x_1}{3} = \frac{2}{9}$, also in the interval $(0,1)$.
• Inductive step: Assume that for some integer $k \geq 1$, both $x_{2k-1}$ and $x_{2k}$ satisfy $0 < x_{2k-1} < 1$ and $0 < x_{2k} < 1$. Now we verify the next two terms:
  • $x_{2k+1} = \frac{x_{2k}}{3} + \frac{2}{3}$. Since $0 < x_{2k} < 1$, substituting gives $0 + \frac{2}{3} < x_{2k+1} < \frac{1}{3} + \frac{2}{3} = 1$, so $x_{2k+1} \in (0,1)$.
  • $x_{2k+2} = \frac{x_{2k+1}}{3}$. Since $0 < x_{2k+1} < 1$, dividing by 3 keeps it in $(0, \frac{1}{3})$, which is a subset of $(0,1)$.

By induction, every term $x_n$ satisfies $0 < x_n < 1$—boundedness is proven!

2. Prove the sequence does not converge #

Your split into odd and even subsequences is exactly the right approach here. Let's formalize each subsequence's limit, then use that to show the original sequence can't converge.

Odd-indexed subsequence: $y_n = x_{2n+1}$ #

You derived the recurrence $y_n = \frac{y_{n-1}}{9} + \frac{2}{3}$ (since $x_{2n+1} = \frac{x_{2n}}{3} + \frac{2}{3} = \frac{x_{2n-1}/3}{3} + \frac{2}{3} = \frac{x_{2n-1}}{9} + \frac{2}{3}$, so $y_n = \frac{y_{n-1}}{9} + \frac{2}{3}$).

This is a linear nonhomogeneous recurrence. To find its limit, we solve for the fixed point $L$ (where $y_n$ stabilizes):
$$
L = \frac{L}{9} + \frac{2}{3}
$$
Rearranging gives:
$$
L - \frac{L}{9} = \frac{2}{3} \implies \frac{8L}{9} = \frac{2}{3} \implies L = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4}
$$

To confirm $y_n \to \frac{3}{4}$, look at the difference between $y_n$ and $L$:
$$
|y_n - \frac{3}{4}| = \left| \frac{y_{n-1}}{9} + \frac{2}{3} - \frac{3}{4} \right| = \left| \frac{y_{n-1}}{9} - \frac{1}{12} \right| = \frac{1}{9} \left| y_{n-1} - \frac{3}{4} \right|
$$
This is a geometric sequence with common ratio $\frac{1}{9} < 1$, so as $n \to \infty$, the difference shrinks to 0. Thus $y_n \to \frac{3}{4}$.

Even-indexed subsequence: $z_n = x_{2n}$ #

Your recurrence here is $z_n = \frac{z_{n-1}}{9} + \frac{2}{9}$ (since $x_{2n} = \frac{x_{2n-1}}{3} = \frac{(\frac{x_{2n-2}}{3} + \frac{2}{3})}{3} = \frac{x_{2n-2}}{9} + \frac{2}{9}$, so $z_n = \frac{z_{n-1}}{9} + \frac{2}{9}$).

Find the fixed point $M$ for this subsequence:
$$
M = \frac{M}{9} + \frac{2}{9}
$$
Rearranging:
$$
\frac{8M}{9} = \frac{2}{9} \implies M = \frac{1}{4}
$$

Check the difference between $z_n$ and $M$:
$$
|z_n - \frac{1}{4}| = \left| \frac{z_{n-1}}{9} + \frac{2}{9} - \frac{1}{4} \right| = \left| \frac{z_{n-1}}{9} - \frac{1}{36} \right| = \frac{1}{9} \left| z_{n-1} - \frac{1}{4} \right|
$$
Again, this is a geometric sequence with ratio $\frac{1}{9} < 1$, so $|z_n - \frac{1}{4}| \to 0$, meaning $z_n \to \frac{1}{4}$.

Final conclusion for non-convergence #

A fundamental property of convergent sequences is that all its subsequences must converge to the same limit. But here, our odd subsequence converges to $\frac{3}{4}$ and the even subsequence converges to $\frac{1}{4}$—two distinct values. This contradicts the requirement for convergence, so the original sequence ${x_n}$ cannot converge.

内容的提问来源于stack exchange，提问作者Alderson