请求证明：当u₀≠0且q<-1时，数列uₙ=u₀qⁿ发散

Hey there! Let's work through this proof step by step—no need to feel stuck anymore. First, let's recap key definitions to ground ourselves:

• A sequence converges if there exists a finite number (L) such that for any small positive (\varepsilon), all terms of the sequence beyond some index (N) lie within (\varepsilon) of (L).
• A sequence diverges if it does NOT converge—either it grows without bound, oscillates without settling to a limit, etc.

Step 1: Analyze the absolute value of the sequence terms #

Given (q < -1), we know (|q| > 1). Let's compute the absolute value of (u_n):
[|u_n| = |u_0 q^n| = |u_0| \cdot |q|^n]
Since (|u_0| > 0) (because (u_0 \neq 0)) and (|q| > 1), the term (|q|^n) grows infinitely large as (n) approaches infinity. That means (|u_n| \to \infty) as (n \to \infty).

Step 2: Use the boundedness property of convergent sequences #

A fundamental theorem about convergent sequences says: If a sequence converges, it must be bounded (i.e., there exists some finite number (M) such that (|u_n| \leq M) for all (n)).

But we just showed (|u_n|) grows without bound—there's no such finite (M) that can contain all terms of the sequence. Therefore, the sequence cannot converge.

Step 3: Add intuition about alternating signs #

Since (q < -1), the sequence terms alternate between positive and negative:

• When (n) is even: (q^n) is positive, so (u_n = u_0 \cdot \text{positive})
• When (n) is odd: (q^n) is negative, so (u_n = u_0 \cdot \text{negative})

Not only do the terms get larger in absolute value, they flip sign every time. This means the sequence bounces between increasingly large positive and negative values—there's no way it can settle down to a single finite limit.

Formal proof by contradiction (optional, for rigor) #

Suppose for contradiction that ({u_n}) converges to a finite limit (L). Then for any (\varepsilon > 0), there exists an integer (N) such that for all (n > N):
[|u_n - L| < \varepsilon]
But (u_{n+1} = q u_n), so substituting into the convergence condition:
[|q u_n - L| < \varepsilon]
But as (n \to \infty), (|u_n| \to \infty), so (|q u_n| = |q| |u_n| \to \infty). This means (|q u_n - L|) will eventually be larger than any fixed (\varepsilon), which contradicts our assumption that the sequence converges. Thus, the sequence must diverge.

内容的提问来源于stack exchange，提问作者QuentinJanuel