Contact: Thorsten Neuschel
thorsten.neuschel@dcu.ie
X133
Course structure:
Assessment schedule:
100% Continuous Assessment:
Final grade = \(\max \{ 0.1 \text{ workshops} + 0.15 \text{ Prelim.} + 0.75 \text{ Final}, 0.9 \text{ Final} \}\).
Weekly routine:
We begin by fixing some notation.
\(\mathbb{N} = \{1, 2, 3, \dots\} \quad \text{"natural numbers"}\)
\(\mathbb{N}_0 = \mathbb{N} \cup \{0\}\)
\(\mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, 3, \dots\} \quad \text{"integers"}\)
\(\mathbb{Q} = \{ \frac{p}{q} : p \in \mathbb{Z}, q \in \mathbb{N} \} \quad \text{"rational numbers"}\)
\(\mathbb{R} = \text{completion of } \mathbb{Q} \quad \text{"real numbers"}\)
Definition. A sequence of real numbers \((a_n)_{n=1}^\infty\) is a function \(f: \mathbb{N} \to \mathbb{R}\), \(f(n) = a_n\). More generally, for any \(k \in \mathbb{Z}\), a sequence \((a_n)_{n=k}^\infty\) is a function \(f: \{k, k+1, \dots\} \to \mathbb{R}\), \(f(n) = a_n\).
Examples. i) The function that lists the natural numbers in natural order is given by
\(f: \mathbb{N} \to \mathbb{R}\), \(f(n) = n\).
We can express this function as the sequence \((a_n)_{n=1}^\infty\) with \(a_n = n\) for all \(n \ge 1\), i.e.
\((a_n)_{n=1}^\infty = (a_1, a_2, a_3, \dots) = (1, 2, 3, \dots)\).
ii) Let \(f: \mathbb{N}_0 \to \mathbb{R}\) be the function that gives the value of an investment after \(n\) periods, when the initial investment is \(P_1\), and the interest rate \(r\) is compounded periodically. Then we have
\(f(n) = P(1+r)^n, \quad n \in \mathbb{N}_0\).
It is more natural to express this in terms of a sequence \((P_n)_{n=0}^\infty\) with
\(P_n = P(1+r)^n\).
We note that we have \(P_0 = P(1+r)^0 = P_1\), where we use the convention \(x^0 = 1\) for \(x \in \mathbb{R}\). This example also illustrates the interpretation of the index \(n\) of a sequence \((a_n)_{n=1}^\infty\) as "time".
Notation: In the following, we use the symbol \((a_n)\) to denote a sequence when we do not want to specify the starting index.
We will now introduce one of the most fundamental concepts in analysis.
Definition. A sequence \((a_n)\) is said to be convergent, if there exists a number \(a \in \mathbb{R}\) with the property: for every \(\epsilon > 0\), there exists an integer \(N = N(\epsilon) \in \mathbb{N}\) such that for all \(n \ge N\) we have \(|a_n - a| < \epsilon\).
In this case we say: \(a_n\) converges to a (as \(n \to \infty\)), and \(a\) is the limit of \((a_n)\). We write: \(\lim_{n \to \infty} a_n = a\) or \(a_n \to a\) (\(n \to \infty\)).
A sequence that is not convergent is called divergent.
Definition. We define \(U_\epsilon(a) := (a-\epsilon, a+\epsilon) = \{x \in \mathbb{R} : a-\epsilon < x < a+\epsilon\}\) for \(\epsilon> 0\) and \(a \in \mathbb{R}\), and call it \(\epsilon\)-neighbourhood of the point a. With this we can write
\(\lim_{n \to \infty} a_n = a \iff \forall \epsilon > 0 \exists N \in \mathbb{N} \forall n \ge N: a_n \in U_\epsilon(a)\).
"For every \(\epsilon > 0\), there comes a point in time after which the terms of the sequence stay within the \(\epsilon\)-neighbourhood of the point a."
Theorem 1. A sequence \((a_n)\) has at most one limit.
Proof. We prove the theorem by contradiction.
Assume: \((a_n)\) has two limits \(a, a' \in \mathbb{R}\) with \(a \neq a'\).
We set \(\epsilon := \frac{1}{2}|a-a'|\), then \(\epsilon > 0\) and by the definition of convergence we have
\(\exists N_1 \in \mathbb{N} \quad \forall n \ge N_1 : |a_n - a| < \epsilon\),
\(\exists N_2 \in \mathbb{N} \quad \forall n \ge N_2 : |a_n - a'| < \epsilon\).
Hence, for \(n \ge \max\{N_1, N_2\}\), we have
\(2\epsilon = |a-a'| = |a - a_n + a_n - a'| \le \underbrace{|a-a_n|}_{ <\epsilon} + \underbrace{|a_n-a'|}_{<\epsilon} < 2\epsilon\),
which is a contradiction. It follows that our assumption is wrong, and the statement follows.
Examples. i) We consider the sequence \((a_n)\) defined by \(a_n = \frac{1}{n}\), \(n \in \mathbb{N}\), so \((a_n) = (1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots)\).
We show \(\lim_{n \to \infty} a_n = 0\). Let \(\epsilon > 0\) be a positive number. We want to find \(N \in \mathbb{N}\) such that for all \(n \ge N\): \(|a_n - 0| < \epsilon\).
We have \(|a_n - 0| < \epsilon \iff \frac{1}{n} < \epsilon \iff n> \frac{1}{\epsilon}\).
We choose any \(N \in \mathbb{N}\) with \(N > \frac{1}{\epsilon}\), then we have for all \(n \ge N\):
\(|a_n - 0| = \frac{1}{n} \le \frac{1}{N} < \epsilon\).
ii) We consider the sequence \((a_n)\) defined by \(a_n = (-1)^n\), \(n \ge 0\).
We show that \((a_n)\) is divergent by contradiction. Assume: \((a_n)\) has a limit \(a \in \mathbb{R}\). We set \(\epsilon := \max\{|1-a|, |-1-a|\} > 0\). Then \(|a_n-a| = |(-1)^n - a| = \epsilon\) for an infinite number of indices \(n\). Hence, for every \(N \in \mathbb{N}\) we find an index \(n \ge N\) with \(|a_n - a| = \epsilon\), which shows that \(a\) cannot be the limit of \((a_n)\). (contradiction).
Definition. A sequence \((a_n)\) is bounded above if there exists a constant \(K \in \mathbb{R}\) such that \(a_n \le K\) for all indices \(n\).
A sequence \((a_n)\) is bounded below if there exists a constant \(K \in \mathbb{R}\) such that \(a_n \ge K\) for all indices \(n\).
Moreover, a sequence \((a_n)\) is bounded if there exists a constant \(K \in \mathbb{R}\) such that \(|a_n| \le K\) for all indices \(n\).
Remark. If \((a_n)\) and \((b_n)\) are bounded sequences, then the sequences \((a_n+b_n)\) and \((a_n b_n)\) are also bounded.
Theorem 2. Every convergent sequence is bounded.
Proof. Let \((a_n)\) be a convergent sequence. W.l.o.g. (without loss of generality), we may assume \((a_n) = (a_n)_{n=1}^\infty\) and that \(\lim_{n \to \infty} a_n = a\).
\(\implies \exists N \in \mathbb{N} \quad \forall n \ge N : |a_n - a| < 1\).
\(\implies \forall n \ge N : |a_n| = |a_n - a + a| \le |a_n - a| + |a| < 1 + |a|\).
Hence, we obtain for all \(n \ge 1\):
\(|a_n| \le \max\{|a_1|, |a_2|, \dots, |a_{N-1}|, 1+|a|\}\).
Remark. The converse of Thm 2 is not true: consider \((a_n)\) given by \(a_n = (-1)^n\), \(n \ge 0\), then obviously \(|a_n| = |(-1)^n| \le 1\) for all \(n \ge 0\), but \((a_n)\) is not convergent.
Theorem 3. Let \((a_n)\) and \((b_n)\) be convergent sequences with limits \(a\) and \(b\), respectively.
Then we have
i) \((|a_n|)\) is convergent with limit \(|a|\),
ii) \((a_n+b_n)\) is convergent with limit \(a+b\),
iii) \((a_n b_n)\) is convergent with limit \(ab\),
iv) if \(b \neq 0\), then \((c_n)\) with \(c_n := \begin{cases} \frac{a_n}{b_n}, & \text{if } b_n \neq 0 \\ 0, & \text{if } b_n = 0 \end{cases}\) is convergent with limit \(\frac{a}{b}\).
Proof. W.l.o.g. we assume \((a_n) = (a_n)_{n=1}^\infty\) and \((b_n) = (b_n)_{n=1}^\infty\). Let \(\epsilon > 0\).
i) We know \(a_n \to a\), so we know \(\exists N_1 \in \mathbb{N} \forall n \ge N_1 : |a_n-a| < \epsilon\).
\(\implies \text{(reverse triangle inequality)} \quad \forall n \ge N_1 : ||a_n| - |a|| \le |a_n-a| < \epsilon\).
ii) We know \(a_n \to a\), \(b_n \to b\), so we know
\(\exists N_1 \in \mathbb{N} \quad \forall n \ge N_1 : |a_n-a| < \epsilon/2\)
\(\exists N_2 \in \mathbb{N} \quad \forall n \ge N_2 : |b_n-b| < \epsilon/2\)
\(\implies \forall n \ge \max\{N_1, N_2\} : |a_n+b_n - (a+b)| = |a_n-a + b_n-b| \le \underbrace{|a_n-a|}_{<\epsilon /2} + \underbrace{|b_n-b|}_{<\epsilon/2} < \epsilon\).
iii) We know \(b_n \to b \implies (\text{Thm 2}) \exists K > 0 : |b_n| \le K\) for all \(n \ge 1\).
Moreover, from \(b_n \to b\) we know
\(\exists N_1 \in \mathbb{N} \quad \forall n \ge N_1 : |a|\,|b_n-b| < \epsilon/2\).
Furthermore, \(a_n \to a\), so we know
\(\exists N_2 \in \mathbb{N} \quad \forall n \ge N_2 : |a_n-a| < \frac{\epsilon}{2K}\).
\[ \begin{aligned} \forall n \ge \max\{N_1, N_2\}:\quad |a_n b_n - ab| &= |(a_n-a)b_n + a(b_n-b)| \\ &\le |a_n-a|\,|b_n| + |a|\,|b_n-b| \\ &\le K|a_n-a| + |a|\,|b_n-b| \\ &< \epsilon/2 + \epsilon/2=\epsilon. \end{aligned} \]
iv) We know \(b \neq 0\), and by part i) we know \(|b_n| \to |b|\).
\(\implies \exists N_1 \in \mathbb{N} \quad \forall n \ge N_1 : |b_n| \ge \frac{|b|}{2}\).
Moreover, since \(a_n \to a\), we know
\(\exists N_2 \in \mathbb{N} \quad \forall n \ge N_2 : \frac{2}{|b|}|a_n-a| < \epsilon/2\),
and as \(b_n \to b\), we know
\(\exists N_3 \in \mathbb{N} \quad \forall n \ge N_3 : \frac{2|a|}{|b|^2}|b_n-b| < \epsilon/2\).
\[ \begin{aligned} \forall n \ge \max\{N_1, N_2, N_3\}:\quad \left|\frac{a_n}{b_n} - \frac{a}{b}\right| &= \left|\frac{a_n b - a b_n}{b b_n}\right| \\ &\le \frac{2}{|b|^2}|a_n b - a b_n| \\ &= \frac{2}{|b|^2}|(a_n-a)b + a(b-b_n)| \\ &\le \frac{2}{|b|}|a_n-a| + \frac{2|a|}{|b|^2}|b-b_n| \\ &< \epsilon/2 + \epsilon/2=\epsilon. \end{aligned} \]