How to make the web version of GitHub Copilot use the correct inline formula version when outputting mathematical formulas?

I found that even after using prompts, inline formulas still cannot be displayed correctly when using the web version of GitHub Copilot.

Some inline formulas are correctly wrapped in a single $ so that the formulas are displayed correctly, but most inline formulas are incorrectly wrapped in two $s, making the formulas unable to display correctly.

The link to the web version of GitHub Copilot is https://github.com/copilot

The model version I am using is o3-mini.

What prompt should I use to display the formula correctly, or is it an internal issue with GitHub Copilot Web?

Below is the personal instruction I used.

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

Below is an example. You can see that some inline formulas are incorrectly enclosed with two dollar signs, so they are not displayed correctly.

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.

I have revised your prompt, it displays correctly the format, but you may encounter some errors during streaming - you may refresh F5 the page to have good looks. Both GPT-o3-mini and Claude 3.7 Sonnet Thinking support good

After refreshing the page

Prompt

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"

Result

# 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. ■