为何当0<t<1、εᵢ=±1时，乘积极限limₙ→∞∏ᵢ=1ⁿ(1+εᵢt)仅能为0？

Great question! Let's unpack this carefully. First, let's formalize the setup: we have a product sequence Pₙ = ∏ᵢ=1ⁿ (1+εᵢt) where 0 < t < 1 and each εᵢ is either 1 or -1. We're told if limₙ→∞ Pₙ exists, it can only be 0—here's why:

Step 1: Assume the limit is non-zero (and reach a contradiction) #

Suppose for contradiction that limₙ→∞ Pₙ = L ≠ 0. Since every factor 1+εᵢt is positive (because 0 < t < 1, so 1+t > 0 and 1-t > 0), Pₙ is always positive, so L > 0.

Step 2: Analyze the ratio of consecutive terms #

Consider the ratio Pₙ₊₁ / Pₙ = 1+εₙ₊₁t. If Pₙ converges to L ≠ 0, then the limit of this ratio must be:
$$\lim_{n→∞} \frac{P_{n+1}}{P_n} = \frac{\lim_{n→∞} P_{n+1}}{\lim_{n→∞} P_n} = \frac{L}{L} = 1$$
But here's the critical issue: this ratio can only take two fixed values, neither of which equals 1:

• When εₙ₊₁ = 1, the ratio is 1+t > 1 (since t > 0)
• When εₙ₊₁ = -1, the ratio is 1-t < 1 (since t > 0)

A sequence made up entirely of two distinct non-1 values can never converge to 1. This directly contradicts our assumption that L ≠ 0.

Step 3: Check pattern-based choices of εᵢ #

Even if we pick εᵢ in structured patterns (alternating 1 and -1, blocks of 1s followed by blocks of -1s), the result still holds:

• Take the logarithm of Pₙ: ln(Pₙ) = ∑ᵢ=1ⁿ ln(1+εᵢt). Let a = ln(1+t) (positive) and b = ln(1-t) (negative, since 1-t < 1).
• The sum ∑ᵢ=1ⁿ ln(1+εᵢt) is a sum of fixed non-zero terms. For this sum to converge to a finite value (required if Pₙ converges to L > 0), the individual terms would need to approach 0—which they never do. No matter how we alternate or group as and bs, the sum will either blow up to +∞ (if we have infinitely more as, making Pₙ→∞) or dive to -∞ (even with equal counts, since a + b = ln(1-t²) < 0, making ln(Pₙ)→-∞ and Pₙ→0).

Putting it all together: any non-zero limit leads to a logical contradiction, so the only possible existing limit is 0.

内容的提问来源于stack exchange，提问作者xyz