Definition – Sequence – Math-reference.net

Definition

A sequence is any function whose domain is $\mathbb{N}$ . Depending on which values of $\mathbb{N}$ the function is actually defined for, the range of the function may either be finite or infinite. If the set of elements of $\mathbb{N}$ that the sequence takes as input values is infinite, then the range will be infinite also. If the number of elements that the sequence accepts is finite, then so will the range be finite also.

Notation

If a sequence $s$ has finite range, then we denote the range of the sequence by $(s_n)_{n=m_1}^{m_2}$ . Ie, $(s_n)_{n=m_1}^{m_2} = s_{m_1}, s_{m_1 +1}, ..., s_{m_2}$ , where $m_1, m_2 \in \mathbb{N}$ . If the sequence is infinite, we denote its range by $(s_n)_{n=1}^{\infty}$ .

Regardless of whether we are dealing with a finite or infinite sequence, we write $s_i = t$ to denote that $(i, t) \in s$ , where $(i, t)\in s \Leftrightarrow s(i) = t$ .

Given a sequence $(s_n)$ we say that $(s_n)$ is increasing if $s_n \leq s_{n+1}$ for all $n\in\mathbb{N}$ . If $s_n \geq s_{n+1}$ for all $n\in\mathbb{N}$ , then we say that $(s_n)$ is decreasing. If $(s_n)$ us either increasing nor decreasing, we say that $(s_n)$ is monotone. Thus, if a sequence is monotone iff it is decreasing or increasing (but not both).

You may also find the terms decreasing and increasing reserved for sequences such that $s_n > s_{n+1}$ and $s_n < s_{n+1}$ respectively.

If a sequence $(s_n)$ is such that there exists an $N\in\mathbb{N}$ such that for all $n,m>N$ ,

$|s_n - s_m| < \epsilon$

then we refer to $(s_n)$ as a Cauchy sequence.

Theorem – Every Cauchy sequence is a convergent sequence. Since every convergent sequence is bounded, we have that every Cauchy sequence is bounded.

We will prove that every Cauchy sequence is convergent. Firstly, we have from the definition of a Cauchy sequence, for every $\epsilon > 0$ , there exists an $N\in\mathbb{R}$ such that $|s_n - s_m| < \epsilon$ for all $n>N$ . We now make two applications of the triangle inequality. Before we do so, note that

$|s_n - s_m| = |s_n - s_m + L - L| \\ = |(s_n - L) - (s_m - L)| = |(s_n - L) + (L - s_m)|$ .

Firstly, we have $|s_n - L| - |s_m - L| \leq |(s_n - L) - (s_m - L)| < \epsilon$ which implies that $|s_n - L| - |s_m - L| < \epsilon$ .

Secondly, $|(s_n - L) + (L - s_m)| < \epsilon$ and $|(s_n - L) + (L - s_m)| \leq |s_n - L| + |L-s_m| = |s_n - L| + |s_m - L|$ , which is equivalent to, $|(s_n - L) + (L - s_m)| \leq |s_n - L| + |s_m - L|$ . Both of these inequalities combine to give $|s_n - L| + |s_m - L| \leq \epsilon$

Thus, we have two inequalities,

$|s_n - L| - |s_m - L| < \epsilon \quad (1)\\\\ |s_n - L| + |s_m - L| \leq \epsilon \quad (2)$

So, adding $(2)$ to equation $(1)$ , we obtain,

$2|s_n - L| < 2\epsilon \Leftrightarrow |s_n - L| < \epsilon \quad (3)$ .

Recall that this is a consequence of the assumption, which was that for every $\epsilon > 0$ , there exists $m,n\in\mathbb{N}$ , such that the hypothesis is fulfilled. But we derived $(3)$ from the hypothesis, so the theorem is proven, and $lim~s_n = L$ .