判断递归序列$a_{n+1}=3-\frac{1}{a_n}(a_1=1)$的敛散性

Hey folks, let's dig into whether this recursive sequence converges. Here's my step-by-step breakdown:

1. 用数学归纳法证明序列正且严格单调递增 #

基例验证 #

First, check the initial terms:

• $a_1 = 1$
• $a_2 = 3 - \frac{1}{a_1} = 3 - 1 = 2$

Clearly, $a_1 < a_2$ and both terms are positive. That's our base case sorted.

归纳假设 #

Suppose for some positive integer $n$, we have $0 < a_n < a_{n+1}$.

归纳步骤 #

We need to show this implies $a_{n+1} < a_{n+2}$ and $a_{n+2} > 0$.

先证正性

From the inductive hypothesis, $a_{n+1} > 0$. Since $a_{n+1} > a_n \geq 1$, we know $\frac{1}{a_{n+1}} < 1$. So:
$$a_{n+2} = 3 - \frac{1}{a_{n+1}} > 3 - 1 = 2 > 0$$
Perfect, $a_{n+2}$ is positive.

再证单调性

Calculate the difference between consecutive terms:
$$
\begin{align*}
a_{n+2} - a_{n+1} &= \left(3 - \frac{1}{a_{n+1}}\right) - \left(3 - \frac{1}{a_n}\right) \
&= \frac{1}{a_n} - \frac{1}{a_{n+1}} \
&= \frac{a_{n+1} - a_n}{a_n a_{n+1}}
\end{align*}
$$
By our inductive hypothesis, $a_{n+1} - a_n > 0$, and $a_n, a_{n+1} > 0$. So the entire fraction is positive, meaning $a_{n+2} - a_{n+1} > 0$, or $a_{n+1} < a_{n+2}$.

By induction, the sequence is strictly increasing and always positive for all $n \geq 1$.

2. 证明序列有上界 #

A strictly increasing sequence converges if it's bounded above. Let's prove our sequence has an upper bound (we can use the eventual limit as a tight upper bound):

First, solve for the potential limit $L$ by assuming $\lim_{n\to\infty}a_n = L$. Plug into the recurrence:
$$L = 3 - \frac{1}{L}$$
Multiply through by $L$:
$$L^2 - 3L + 1 = 0$$
Solving this quadratic gives roots $\frac{3+\sqrt{5}}{2} \approx 2.618$ and $\frac{3-\sqrt{5}}{2} \approx 0.382$. Since our sequence starts at 1 and increases, we'll focus on the larger root as our upper bound.

归纳法证明 $a_n < \frac{3+\sqrt{5}}{2}$ for all $n$ #

• Base case: $a_1=1 < 2.618$, $a_2=2 < 2.618$ — holds.
• Inductive hypothesis: Assume $a_n < \frac{3+\sqrt{5}}{2}$.
• Inductive step: We need to show $a_{n+1} < \frac{3+\sqrt{5}}{2}$.

Start with the recurrence:
$$a_{n+1} = 3 - \frac{1}{a_n}$$
We need to show:
$$3 - \frac{1}{a_n} < \frac{3+\sqrt{5}}{2}$$
Rearrange terms:
$$3 - \frac{3+\sqrt{5}}{2} < \frac{1}{a_n}$$
Calculate the left-hand side:
$$\frac{6 - 3 - \sqrt{5}}{2} = \frac{3-\sqrt{5}}{2}$$
Note that $\frac{1}{\frac{3+\sqrt{5}}{2}} = \frac{2}{3+\sqrt{5}} = \frac{3-\sqrt{5}}{2}$ (rationalize the denominator to confirm). So our inequality becomes:
$$\frac{3-\sqrt{5}}{2} < \frac{1}{a_n} \implies a_n < \frac{3+\sqrt{5}}{2}$$
Which is exactly our inductive hypothesis. So the upper bound holds for all $n$.

3. 结论：序列收敛 #

Since the sequence is strictly increasing and bounded above, by the Monotone Convergence Theorem (for real numbers), the sequence must converge. Its limit is $\frac{3+\sqrt{5}}{2}$, as we derived earlier.

内容的提问来源于stack exchange，提问作者TheValars