我发现即使使用了提示词，在使用 GitHub Copilot 的网页版时，行内公式仍然无法正确显示。

一些行内公式被正确地包裹在单个 $ 中，因此公式可以正确显示，但大多数行内公式都被错误地包裹在两个 $$ 中，导致公式无法正确显示。

GitHub Copilot 网页版的链接是 https://github.com/copilot

我使用的模型版本是 o3-mini。

我应该使用什么提示词才能正确显示公式，或者这是否是 GitHub Copilot Web 的内部问题？

下面是我使用的个人指令。

Please use the correct LaTeX math formula formatting in your responses:

1. Inline formulas: Use single dollar signs, for example $E=mc^2$

2. Display formulas: Use double dollar signs

下面是一个例子。你可以看到一些行内公式被错误地用两个美元符号括起来，所以它们没有被正确显示。

The Leibniz criterion (or alternating series test) states that an alternating series

$$
\sum_{n=1}^{\infty} (-1)^{n-1} a_n,
$$

converges if the sequence $(a_n)$ satisfies the following two conditions:

1. The sequence is monotonically decreasing: 
   $$
   a_1 \ge a_2 \ge a_3 \ge \cdots \ge 0,
   $$
2. The terms converge to zero:
   $$
   \lim_{n\to\infty} a_n = 0.
   $$

Below is a classical proof of the criterion.

---

**Proof:**

Define the \( n \)th partial sum of the series by

$$
s_n = \sum_{k=1}^n (-1)^{k-1} a_k.
$$

We will show that the sequence \( (s_n) \) converges by demonstrating that the partial sums form two sequences—one consisting of the even-indexed partial sums and one consisting of the odd-indexed partial sums—that are monotonic and bounded, and that they “squeeze” the series into a unique limit.

1. **Consider the subsequence of odd-indexed partial sums:**

   Let
   $$
   s_{2n-1} = a_1 - a_2 + a_3 - \cdots + a_{2n-1}.
   $$
   
   To see that \( (s_{2n-1}) \) is decreasing, observe that

   $$
   s_{2n-1} - s_{2n+1} = \Big[a_1 - a_2 + \cdots + a_{2n-1}\Big] - \Big[a_1 - a_2 + \cdots + a_{2n-1} - a_{2n} + a_{2n+1}\Big].
   $$
   
   This simplifies to

   $$
   s_{2n-1} - s_{2n+1} = a_{2n} - a_{2n+1}.
   $$
   
   Since \( a_{2n} \ge a_{2n+1} \) (by the monotonicity assumption), we have

   $$
   s_{2n-1} \ge s_{2n+1}.
   $$

   Thus, the odd-indexed partial sums form a decreasing sequence.

2. **Consider the subsequence of even-indexed partial sums:**

   Let
   $$
   s_{2n} = a_1 - a_2 + a_3 - \cdots - a_{2n}.
   $$
   
   Now, compare consecutive even-indexed partial sums:

   $$
   s_{2n+2} - s_{2n} = \Big[s_{2n} - a_{2n+1} + a_{2n+2}\Big] - s_{2n} = - a_{2n+1} + a_{2n+2}.
   $$
   
   Because \( a_{2n+1} \ge a_{2n+2} \), it follows that

   $$
   - a_{2n+1} + a_{2n+2} \ge 0,
   $$
   
   so

   $$
   s_{2n+2} \ge s_{2n}.
   $$
   
   Therefore, the even-indexed partial sums form an increasing sequence.

3. **Bounding the two subsequences:**

   Notice that for every \( n \),
   $$
   s_{2n} \le s_{2n-1}.
   $$
   
   This is because \( s_{2n-1} = s_{2n} + a_{2n} \) and \( a_{2n} \ge 0 \).

   Consequently, the entire sequence of partial sums is "squeezed" between these two subsequences:
   
   - The increasing sequence \( (s_{2n}) \) is bounded above by \( s_1 \) (since \( s_1 \) is the first odd partial sum).
   - The decreasing sequence \( (s_{2n-1}) \) is bounded below by \( s_{2n} \).

   By the Monotone Convergence Theorem, both sequences converge; denote

   $$
   \lim_{n\to\infty} s_{2n} = L \quad \text{and} \quad \lim_{n\to\infty} s_{2n-1} = M.
   $$
   
   Since
   $$
   s_{2n} \le s_{2n-1}
   $$
   for all \( n \) and the difference between them is \( s_{2n-1} - s_{2n} = a_{2n} \), which tends to 0 by assumption, we obtain
   $$
   \lim_{n\to\infty} \big(s_{2n-1} - s_{2n}\big) = 0.
   $$
   Therefore, \( L = M \).

4. **Conclusion:**

   Since both subsequences converge to the same limit \( L \), the sequence of partial sums \( (s_n) \) converges to \( L \). That is,

   $$
   \sum_{n=1}^{\infty} (-1)^{n-1} a_n = L.
   $$

Thus, the Leibniz criterion confirms that the alternating series converges.

---

This completes the proof.

回答：

我修改了你的提示词，它现在能正确显示格式了，但你可能会在流式传输过程中遇到一些错误 – 你可以刷新页面（F5）来获得良好的效果。 GPT-o3-mini 和 Claude 3.7 Sonnet Thinking 都支持得很好。

刷新页面后

提示词

Please use the correct **LaTeX math formula formatting in Mardown** in your responses:
1. Inline formulas: Use single dollar signs, for example $E=mc^2$
2. Display formulas: Use double dollar signs with lines break, e.g.

$$
S_{2n+1} - S_{2n} = a_{2n+1}
$$

3. Handle well error "extra open brace or missing close brace"

结果

# Proof of the Leibniz Criterion for Alternating Series

The Leibniz criterion provides conditions for the convergence of alternating series of the form $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$ or $\sum_{n=1}^{\infty} (-1)^{n} a_n$.

## Theorem (Leibniz Criterion)

Let $\{a_n\}$ be a sequence satisfying:
1. $a_n > 0$ for all $n \geq 1$
2. $a_n \geq a_{n+1}$ for all $n \geq 1$ (monotonically decreasing)
3. $\lim_{n \to \infty} a_n = 0$

Then the alternating series $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$ converges.

## Proof

Let's define the partial sums:

$$
S_n = \sum_{k=1}^{n} (-1)^{k+1} a_k
$$

We'll examine the behavior of even-indexed partial sums $S_{2n}$ and odd-indexed partial sums $S_{2n+1}$.

### Step 1: Show even-indexed partial sums form an increasing sequence

For consecutive even-indexed partial sums:

$$
S_{2n+2} - S_{2n} = (-1)^{2n+2+1}a_{2n+1} + (-1)^{2n+3+1}a_{2n+2} = a_{2n+1} - a_{2n+2}
$$

Since $a_{2n+1} \geq a_{2n+2}$ (by monotonicity), we have $S_{2n+2} - S_{2n} \geq 0$, meaning:

$$
S_2 \leq S_4 \leq S_6 \leq \ldots \leq S_{2n} \leq S_{2n+2} \leq \ldots
$$

### Step 2: Show odd-indexed partial sums form a decreasing sequence

Similarly, for consecutive odd-indexed partial sums:

$$
S_{2n+1} - S_{2n-1} = (-1)^{2n+1}a_{2n} + (-1)^{2n+2}a_{2n+1} = -a_{2n} + a_{2n+1}
$$

Since $a_{2n} \geq a_{2n+1}$ (by monotonicity), we have $S_{2n+1} - S_{2n-1} \leq 0$, meaning:

$$
S_1 \geq S_3 \geq S_5 \geq \ldots \geq S_{2n-1} \geq S_{2n+1} \geq \ldots
$$

### Step 3: Show even-indexed partial sums are bounded above

For any $n \geq 1$:

$$
\begin{align*}
S_{2n} &= a_1 - a_2 + a_3 - \ldots + a_{2n-1} - a_{2n} \\
&= a_1 - (a_2 - a_3) - (a_4 - a_5) - \ldots - (a_{2n-2} - a_{2n-1}) - a_{2n}
\end{align*}
$$

Since $a_k \geq a_{k+1}$ for all $k$, each term $(a_{2k} - a_{2k+1}) \geq 0$, so:

$$
S_{2n} \leq a_1
$$

### Step 4: Show odd-indexed partial sums are bounded below

Similarly, for odd-indexed partial sums:

$$
\begin{align*}
S_{2n+1} &= a_1 - a_2 + a_3 - \ldots + a_{2n-1} - a_{2n} + a_{2n+1} \\
&= a_1 - (a_2 - a_3) - (a_4 - a_5) - \ldots - (a_{2n} - a_{2n+1})
\end{align*}
$$

Each term $(a_{2k} - a_{2k+1}) \geq 0$, so:

$$
S_{2n+1} \geq a_1 - (a_2 - 0) = a_1 - a_2
$$

### Step 5: Establish convergence 

Since $\{S_{2n}\}$ is increasing and bounded above, it converges to some limit $L_1$.
Since $\{S_{2n+1}\}$ is decreasing and bounded below, it converges to some limit $L_2$.

Now, we need to show $L_1 = L_2$. Observe:

$$
S_{2n+1} - S_{2n} = a_{2n+1}
$$

Taking the limit as $n \to \infty$ and using the condition $\lim_{n \to \infty} a_n = 0$:

$$
\lim_{n \to \infty} (S_{2n+1} - S_{2n}) = \lim_{n \to \infty} a_{2n+1} = 0
$$

Therefore, $L_2 - L_1 = 0$, which means $L_1 = L_2 = L$.

Since both subsequences of partial sums converge to the same limit $L$, the entire sequence $\{S_n\}$ converges to $L$, proving that the alternating series $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$ converges.

### Error Bound

An additional important result is that if we approximate the sum by a partial sum $S_n$, the error is bounded by the next term:

$$
|S - S_n| \leq a_{n+1}
$$

Where $S$ is the sum of the infinite series.

This completes the proof of the Leibniz criterion. ■