Contractive sequence: $|x_{k+2} - x_{k+1}| \le c\,|x_{k+1} - x_k|$ for a fixed $0 < c < 1$

A sequence $(x_k)$ of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) is contractive when there is a real $c$ with

$0 < c < 1 \qquad \text{and} \qquad |x_{k+2} - x_{k+1}| \;\le\; c\,|x_{k+1} - x_k| \quad \text{for every } k \in \mathbb{N},$

the order and the absolute value being those of $\mathbb{R}$ (Order on the reals, Basic properties of the absolute value). Such a $c$ is called a contraction constant for $(x_k)$.

The constant must not depend on $k$. This is the whole content of the definition and the only place it can go wrong. A sequence whose consecutive gaps each shrink, so that

$|x_{k+2} - x_{k+1}| < |x_{k+1} - x_k| \quad \text{for every } k,$

is not contractive on that evidence: what is required is a single $c < 1$ working at every index simultaneously. The two conditions really are different: there is a sequence satisfying the second that satisfies the first for no $c < 1$ and does not converge, and it is the named counterexample of the companion page, recalled in the remarks below.

The constant is not unique. If $c$ is a contraction constant then so is every $c'$ with $c \le c' < 1$, since $c\,|x_{k+1} - x_k| \le c'\,|x_{k+1} - x_k|$ when $|x_{k+1} - x_k| \ge 0$ (Basic properties of the absolute value). Statements about contractive sequences therefore quantify over a chosen constant, and the error bound in Every contractive sequence is Cauchy, hence converges, with error bound $|x - x_k| \le c^{k-1}|x_2 - x_1|/(1-c)$ for $k \ge 1$ is sharper for a smaller $c$.

Degenerate cases are included. A constant sequence is contractive with every $c \in (0,1)$, all the gaps being $0$. A sequence that is eventually constant is contractive as soon as the inequality holds at the finitely many earlier indices. Nothing in the definition forces the gaps to be positive.

• Where the name comes from. The typical contractive sequence arises by iterating a map: if $f : \mathbb{R} \to \mathbb{R}$ satisfies $|f(u) - f(v)| \le c\,|u - v|$ for all $u, v$, with $0 < c < 1$, and $x_{k+1} = f(x_k)$, then $(x_k)$ is contractive with the same $c$, because $|x_{k+2} - x_{k+1}| = |f(x_{k+1}) - f(x_k)| \le c\,|x_{k+1} - x_k|$. This is the elementary shadow of the Banach fixed point theorem, and The sequence $x_{k+1} = (x_k + 1)/3$ is contractive with $c = 1/3$ and converges to $1/2$ ↗ is the simplest instance.

• The definition says nothing about a limit, and that is deliberate. It is a condition on consecutive differences only, checkable without knowing where the sequence is going, which is exactly what makes Every contractive sequence is Cauchy, hence converges, with error bound $|x - x_k| \le c^{k-1}|x_2 - x_1|/(1-c)$ for $k \ge 1$ useful: convergence and an explicit error bound both fall out of a hypothesis that never mentions the limit.

• Contractive implies the gaps are null but not conversely. From the definition the gaps satisfy $|x_{k+1} - x_k| \le c^{k-1}|x_2 - x_1|$ for $k \ge 1$, which tends to $0$ (For $|r| < 1$ the sequence $r^k$ is null, and for $|r| > 1$ the sequence $|r|^k$ diverges to $+\infty$); the first gap $|x_1 - x_0|$ is unconstrained, having no predecessor (Every contractive sequence is Cauchy, hence converges, with error bound $|x - x_k| \le c^{k-1}|x_2 - x_1|/(1-c)$ for $k \ge 1$). The converse implication fails badly: gaps tending to $0$ do not even give a Cauchy sequence, which is FALSE: if $|x_{k+1} - x_k| \to 0$ then $(x_k)$ is Cauchy.

• The witness separating the two conditions is $x_{k+1} = x_k + 1/x_k$ from $x_1 = 1$ has strictly decreasing consecutive gaps and diverges, so no uniform $c < 1$ exists ↗: its consecutive gaps strictly decrease, every ratio of consecutive gaps is below $1$, and still no single $c < 1$ works, because those ratios approach $1$. It is what the uniformity requirement above exists to exclude.