Advanced Analysis

Sequences Introduction

Video. Sequences Introduction
Definition
An infinite sequence of real numbers \(\{a_n\}_{n=1}^\infty\) is a function from the natural numbers into the reals. The infinite sequence is said to converge to \(L\) as \(n\) tends to infinity when
\[{}\forall \epsilon > 0, \exists N \in {\mathbb N} \text{ such that }{}\]
\[{}n \geq N \Rightarrow |a_n - L| < \epsilon.{}\]
In words, for all \(\epsilon\) greater than zero, there is some natural number \(N\) such that \(n\) being greater than or equal to \(N\) implies that the distance from \(a_n\) to \(L\) (measured by absolute value of the difference) is strictly smaller than \(\epsilon\). When \(\{a_n\}_{n=1}^\infty\) converges to \(L\) as \(n \rightarrow \infty\), we write
\[ \lim_{n \rightarrow \infty} a_n = L. \]
Lemma
If \(\{b_n\}_{n=1}^\infty\) is a convergent sequence, then the set
\[ \{ x \in {\mathbb R} \ : \ x = b_n \text{ for some } n \in {\mathbb N}\}\]
is bounded, i.e., there exists \(M\) such that \(|b_n| \leq M\) for all \(n \in N\).
Proof
If the limit is \(L\), fix \(\epsilon = 1\); we know that there is some \(N\) such that \(|b_n - L| < 1\) for all \(n \geq N\). The triangle inequality implies that \(|b_n| \leq |L| + 1\) for all such \(b \geq N\). However, the \(b_n\)'s not subject to this inequality are themselves finite, so there is a term \(|b_{n_0}|\) which is equal to the maximum \(\max\{|b_1|,\ldots,|b_{N-1}|\}\) when \(N > 1\) (if \(N =1\), we're in good shape because the inequality \(|b_n| \leq |L| + 1\) applies to all terms.) Thus if
\[ M := \max \{|b_1|,\ldots,|b_{N-1}|,|L|+1\},\]
it follows that \(|b_n| \leq M\) for all \(n \geq 1\).
Theorem (Algebraic Limit Laws)
If \(\lim_{n \rightarrow \infty} a_n = A\) and \(\lim_{n \rightarrow \infty} b_n = B\),
  • \(\lim_{n \rightarrow \infty} (a_n + b_n) = A + B\).
  • \(\lim_{n \rightarrow \infty} (a_n - b_n) = A - B\).
  • \(\lim_{n \rightarrow \infty} a_n b_n = A B\).
  • For any constant \(c\), \(\lim_{n \rightarrow \infty} c a_n = cA\).
  • If \(B \neq 0\), \(\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = \frac{A}{B}\). In particular, assuming that the limits of \(a_n\) and \(b_n\) exist, all the other listed limits must exist as well and have the stated values.
Proof
Meta (Some Ideas)
For the addition law, fix \(\epsilon\) and let \(N_1\) be such that \(|a_n - A| < \epsilon/2\) when \(n \geq N_1\) and \(|b_n-B| < \epsilon/2\) when \(n \geq N_2\). If \(n \geq \max\{N_1,N_2\}\), both inequalities are true, so every such \(n\) satisfies
\[{}|a_n + b_n - A - B| {}\]
\[{}\leq |a_n - A| + |b_n - B|{}\]
\[{}< \epsilon/2 + \epsilon/2 = \epsilon.{}\]
For the ratio law, first observe that since \(B \neq 0\), we may take \(\epsilon = |B|/2\) and we see that \(|b_n - B| < |B|/2\) for all \(n\) greater than or equal to some threshold \(N_0\). By the triangle inequality, it follows that \(|b_n| > |B|/2\) for all such \(n\), so in particular they don't vanish. This is important, because we need the fraction \(a_n/b_n\) to be defined for all but possibly finitely many values of \(n\) if we want to talk about its limit. Now for all such \(n\),
\[{}\left|\frac{a_n}{b_n} - \frac{A}{B}\right|{}\]
\[{}= \left|\frac{B a_n - A b_n}{b_n B}\right|{}\]
\[{}\leq \frac{2}{|B|^2} |B a_n - A b_n|{}\]
Assuming that the other limit laws have already been proved, we know that \(B a_n - A b_n\) converges to \(B A - A B = 0\) when \(n \rightarrow \infty\), so that means that there must be some \(N_1\) for which \(|B a_n - A b_n| < \epsilon |B|^2/2\) for all \(n \geq N_1\). If \(n \geq \max\{N_1,N_2\}\), then both of our inequalities are simultaneously true, and consequently
\[{}\left|\frac{a_n}{b_n} - \frac{A}{B}\right| < \epsilon{}\]
as desired.
Definition
A sequence \(\{a_n\}_{n=1}^\infty\) of real numbers is called a Cauchy sequence when for every \(\epsilon > 0\), there exists some \(N \in {\mathbb N}\) such that \(|a_n - a_m| < \epsilon\) whenever both \(n,m \geq N\).
Meta (How to think about Cauchy Sequences)
Think of \(n\) and \(m\) as being far apart; the only constraint they need to satisfy is that they're both bigger than or equal to \(N\), and there are lots of numbers which are bigger than or equal to \(N\). The indices \(n\) and \(m\) can be extremely far apart for any given \(\epsilon\) and \(N\), and that's the typical situation.
Theorem
A sequence of real numbers \(\{a_n\}_{n=1}^\infty\) is a Cauchy sequence if and only if it converges.
Proof
Postponed until after the Bolzano-Weierstrass Theorem. For the proof, see the upcoming section on Cauchy sequences.
Meta (The Point of Cauchy Sequences)
It's often much easier to prove that a sequence is Cauchy than it is to find the exact value of the limit, which would in principle need to be known in order to prove convergence. Cauchy sequences are almost always behind the scenes when you have a theorem about a sequence whose limit can't be easily guessed from context.