Let ${p_n}$, ${q_n}$ be the sequences that satisfies $p_1 = 2$, $q_1 = 1$, $p_{n+1} = 0.65(p_n + \sqrt{p_n q_n})$, $q_{n+1} = 0.65(p_n - \sqrt{p_n q_n})$ for every $n \in \mathbb{N}$. Then $\lim_{n\to\infty} p_n$ exists and it is 0.

GPT-5.4 (changed mind)

Vote: No

I changed my vote. My round-1 argument contained a numerical error: in fact p_2=0.65(2+\sqrt2)\approx2.219, not 1.919. That already undermines the claimed monotone decay. The decisive analysis is to track the ratio r_n=q_n/p_n. Then r_{n+1}=\frac{1-\sqrt{r_n}}{1+\sqrt{r_n}},\qquad p_{n+1}=0.65(1+\sqrt{r_n})p_n. Starting from r_1=1/2, this ratio map does not drive r_n to 1 or 0 in a way that would force decay of p_n. Gemini 3.1 Pro gave the cleanest killer observation: for this initial value, r_n alternates between two values, so the two-step growth factor is constant: \frac{p_{n+2}}{p_n}=0.65^2\Bigl(1+\frac1{\sqrt2}\Bigr)\bigl(1+(\sqrt2-1)\bigr)=0.65^2(\sqrt2+1)\approx1.020005>1. Therefore p_{n+2}>p_n for all n, with exponential growth along each parity subsequence. So p_n cannot converge to 0; in fact it grows unboundedly. Hence the statement “the limit exists and equals 0” is false.

Claude Opus 4.6

Vote: No

I maintain my vote of B. The key insight, clearly articulated by Gemini 3.1 Pro, is the two-step growth analysis. Let me verify the critical calculation. The ratio x_n = q_n/p_n satisfies x_{n+1} = (1 - √x_n)/(1 + √x_n). With the substitution √x_n = tan(θ_n), we get θ_{n+1} = π/4 - θ_n, so the ratio alternates between two values. Starting with x_1 = 1/2, we get √x_1 = 1/√2 and √x_2 = √2 - 1. The two-step growth factor is: p_{n+2}/p_n = 0.65²(1 + √x_1)(1 + √x_2) = 0.4225 × (1 + 1/√2)(1 + √2 - 1) = 0.4225 × (1 + 1/√2) × √2 = 0.4225 × (√2 + 1) ≈ 0.4225 × 2.4142 ≈ 1.0200. Since this ratio exceeds 1, p_n grows geometrically (roughly by factor 1.02 every two steps), meaning p_n → ∞, not 0. The GPT-5.4 and Mistral Large arguments for A contain errors. Mistral's computation of p_n·q_n is wrong: p_{n+1}·q_{n+1} = 0.65²·p_n(p_n - q_n), NOT 0.65²·p_n·q_n. The product does not decay exponentially. GPT-5.4's numerical values (p_2 ≈ 1.919) are incorrect — the correct value is p_2 ≈ 2.219 as both I and Mistral computed. The fixed point analysis showing (0,0) is the only fixed point is correct but irrelevant — the sequences don't need to converge to a fixed point; they can diverge. And indeed they do, since the per-step multiplicative factor 0.65(1 + √x_n) alternates around a geometric mean of √1.02 > 1. The statement that lim p_n = 0 is false. The answer is B.

Mistral Large (changed mind)

Vote: No

1. **Numerical Evidence Shows Divergence**: The computed terms (e.g., p1=2, p2≈2.219, p3≈2.040, p4≈2.179) oscillate but do not decay to 0. The two-step growth factor (≈1.02 > 1) proves exponential divergence, as shown by Gemini 3.1 Pro and Kimi K2.5. 2. **Fixed-Point Analysis**: The only fixed point is (0,0), but the sequences do not converge to it. The multiplier for p_n (0.65(1 + √r*)) ≈1.0036 > 1, causing p_n → ∞. This aligns with Grok 4.1 Fast and Claude Opus 4.6. 3. **Ratio Dynamics**: The ratio r_n = q_n/p_n converges to r*≈0.2956, but p_n grows because 0.65(1 + √r*) > 1. The trigonometric substitution in Gemini 3.1 Pro confirms oscillatory divergence. 4. **Flaw in Original Argument**: The product p_n q_n → 0 does not imply p_n → 0 if q_n decays faster than p_n grows. Numerical evidence contradicts this assumption. 5. **Consensus**: All models except GPT-5.4 agree the limit is not 0. The combined evidence (numerical, analytical, and ratio-based) is compelling.

Kimi K2.5

Vote: No

The sequence $p_n$ diverges to infinity, so the limit is not 0. Let $x_n = q_n/p_n$. Then $p_{n+1} = 0.65p_n(1+\sqrt{x_n})$ and $x_{n+1} = \frac{1-\sqrt{x_n}}{1+\sqrt{x_n}}$. Starting with $x_1 = 1/2$, the ratio $\sqrt{x_n}$ oscillates between $\sqrt{1/2}$ and $\sqrt{2}-1$. The product of two consecutive growth factors is $0.65^2(1+\sqrt{x_1})(1+\sqrt{x_2}) = 0.65^2(1+1/\sqrt{2})(1+\sqrt{2}-1) = 0.65^2(\sqrt{2}+1) \approx 0.4225 \times 2.414 \approx 1.02 > 1$. Since $p_{n+2}/p_n > 1.02$ for all $n$, the sequence grows exponentially. Numerical verification confirms this: $p_1=2$, $p_2\approx2.22$, $p_3\approx2.04$, $p_4\approx2.18$, showing oscillation with increasing magnitude rather than convergence to 0.

Grok 4.1 Fast

Vote: No

The ratio r_n = q_n / p_n follows r_{n+1} = (1 - \sqrt{r_n}) / (1 + \sqrt{r_n}), which converges to the unique fixed point r^* \approx 0.2956 where \sqrt{r^*} = x^* \approx 0.5437 solves x^3 + x^2 + x - 1 = 0. Numerical iterations confirm oscillation damping toward r^*: r_1=0.5, r_2\approx0.172, r_3\approx0.414, r_4\approx0.217, r_5\approx0.365, r_6\approx0.247, r_7\approx0.336, r_8\approx0.266, approaching r^*. Thus, p_{n+1} / p_n = 0.65 (1 + \sqrt{r_n}) \to 0.65 (1 + x^*) \approx 1.0034 > 1. Eventually, p_n grows geometrically (p_{n+1} > 1.001 p_n), so p_n \to \infty. No finite limit exists, certainly not 0. A voters' claims fail: GPT-5.4 and Mistral Large use incorrect numerics (e.g., p_2\approx1.919 or 2.219 but claim decrease; actual p_n oscillates around 2+ and grows slowly). Mistral's product p_{n+1}q_{n+1} = 0.4225 p_n (p_n - q_n) \approx c p_n^2 (1 - r_n) \to \infty, not decay. Gemini's trig alternation is invalid (r_n doesn't alternate; sqrt(x_{n+1}) \neq tan(\pi/4 - \theta_n) consistently for x_n = tan^2 \theta_n). Claude and Kimi reinforce B with solid numerics/dynamics. Statement false.

Gemini 3.1 Pro

Vote: No

I previously made an error in my trigonometric substitution by confusing $x_{n+1}$ with its square root. I am persuaded by Grok 4.1 Fast's rigorous algebraic approach, which perfectly resolves the asymptotic behavior of the sequence. Let $r_n = rac{q_n}{p_n}$. The recurrence gives $r_{n+1} = rac{1 - sqrt{r_n}}{1 + sqrt{r_n}}$. Substituting $x_n = sqrt{r_n}$, we get $x_{n+1}^2 = rac{1 - x_n}{1 + x_n}$. The sequence $x_n$ is easily shown to converge to $x^*$, the unique positive real root of the equation $x^3 + x^2 + x - 1 = 0$. To determine whether $p_n$ converges to 0, we can analyze the growth factor: $\frac{p_{n+1}}{p_n} = 0.65(1 + \sqrt{r_n}) = 0.65(1 + x_n)$. As $n \to \infty$, this ratio converges to a constant multiplier $k = 0.65(1 + x^*)$. Notice that $0.65 = \frac{13}{20}$. If we evaluate the cubic polynomial $f(x) = x^3 + x^2 + x - 1$ at $x = \frac{7}{13}$, we get: $f(\frac{7}{13}) = (\frac{7}{13})^3 + (\frac{7}{13})^2 + \frac{7}{13} - 1 = \frac{343 + 637 + 1183 - 2197}{2197} = -\frac{34}{2197} < 0$. Since $f(x)$ is strictly increasing for $x > 0$ and evaluates to a negative number at $\frac{7}{13}$, its root must be strictly greater than $\frac{7}{13}$. Thus, $x^* > \frac{7}{13}$. Substituting this back into our limit multiplier: $k = 0.65(1 + x^*) > \frac{13}{20} (1 + \frac{7}{13}) = \frac{13}{20} (\frac{20}{13}) = 1$. Because the asymptotic ratio between consecutive terms of $p_n$ is a constant $k > 1$, the sequence $p_n$ grows exponentially and diverges to $\infty$. The limit does not exist, so the statement is false.

Final Summary

The AI Roundtable initially split over whether the sequence $p_n$ vanished, with GPT-5.4 and Mistral Large falling for the 'illusion of decay' caused by the 0.65 multiplier. However, Gemini 3.1 Pro and Kimi K2.5 dismantled this theory by proving that $p_n$ actually grows exponentially, resulting in a unanimous Round 2 consensus that the limit is not zero.

All 6 models agreed on "No" after discussion

Strongest Arguments

• No: A two-step growth analysis reveals that $p_{n+2}/p_n = 0.65^2(1 + 1/\sqrt{2})(1 + \sqrt{2} - 1) = 0.65^2(\sqrt{2} + 1) \approx 1.02$. Since this ratio is strictly greater than 1, the sequence $p_n$ diverges toward infinity rather than converging to zero.

Key Turning Points

• Gemini 3.1 Pro demonstrated that for the specific initial conditions, the growth factor over two steps is approximately 1.02, proving divergence.
• Grok 4.1 Fast and Kimi K2.5 generalized this by showing the ratio $q_n/p_n$ converges to a fixed point $x^*$, where the resulting multiplier $0.65(1+x^*)$ is roughly 1.0036, confirming long-term exponential growth.
• GPT-5.4 and Mistral Large admitted to numerical and algebraic errors in their first-round reasoning after seeing the growth factor proofs.