Preface
These lecture notes are strongly inspired by the lecture notes of the course “Analysis I” at Technische Universität Berlin by Prof. Dr. Christian Mehl. The provision of these notes by Prof. Mehl is highly appreciated.
You may recognize the material in these notes since most topics that are covered here, have already been discussed in high-school. However, here we will provide precise mathematical definitions and theorems as well as their proofs. In this sense, this course will be much more abstract and also challenging compared to high-school. Here you will not only learn the “How” but also the “Why”. For example, you will not only learn differentiation und integration rules and how to apply them, but also why these rules hold and when a function is actually differentiable and integrable.
https://moodle.fernuni.ch/course/view.php?id=2242
There are many books you can use to learn analysis, you can basically take any of the standard books. If you would like to read some books in German and French we recommend
O. Forster: Analysis 1, 12. Auflage, Springer Spektrum, 2015. ISBN 978-3-658-11544-9.
C. Deschamps, F. Moulin, N. Cleirec, J.-M. Cornil, Y. Gentric, F. Lussier, C. Mullaert, S. Nicolas: Maths MPSI – Tout-en-un, 5\(^{\rm e}\) édition, Dunod, 2018. ISBN 978-2-10-077659-7.
Even though these notes are written with great care, it is not guaranteed that there are no errors (in fact, it is quite likely that there are some). If something is completely unclear or if you find errors or typos, then please be so kind and send an e-mail to matthias.voigt@fernuni.ch. Thank you very much for your help in improving these notes! Your suggestions are very welcome!
Matthias Voigt (Thun, 2024-09-30)
1 Preliminaries: Axioms of Real Numbers
In this short chapter we revisit the real numbers and discuss a few more of their properties that we will make use of in later chapters. It is assumed that you have attended the course M01: “Algorithmics” in which some of the following concepts have already been introduced.
Here we will assume that the set fo real numbers \({\mathbb{R}}\) is given and then we will characterize its properties by so-called axioms (i. e., the basic fundamental statements our entire mathematics is build from). Let us recall some of the axioms of real numbers. First we know that the set of real numbers forms a field.
Axiom 1.1 • Field axioms
On \({\mathbb{R}}\) there exist two operations \[\begin{aligned} +&: {\mathbb{R}}\times {\mathbb{R}}\to {\mathbb{R}}, \quad (a,b) \mapsto a+b, \\ \cdot&: {\mathbb{R}}\times {\mathbb{R}}\to {\mathbb{R}}, \quad (a,b) \mapsto a \cdot b, \end{aligned}\] called addition and multiplication, for which the following properties hold:
Axioms of addition:
Associative law: For all \(a,\,b,\,c \in {\mathbb{R}}\) it holds that \[(a+b)+c = a+(b+c).\]
Commutative law: For all \(a,\,b\in {\mathbb{R}}\) it holds that \[a+b = b+a.\]
Existence of a neutral element: There exists a number \(0 \in {\mathbb{R}},\) called “zero” such that for all \(a \in {\mathbb{R}}\) we have that \[a + 0 = a.\]
Existence of an inverse element: For every \(a \in {\mathbb{R}}\) there exists a number \(\widetilde{a} \in {\mathbb{R}},\) called additive inverse (element) such that \[a + \widetilde{a} = 0.\]
Axioms of multiplication:
Associative law: For all \(a,\,b,\,c \in {\mathbb{R}}\) it holds that \[(a \cdot b) \cdot c = a \cdot (b \cdot c).\]
Commutative law: For all \(a,\,b\in {\mathbb{R}}\) it holds that \[a \cdot b = b \cdot a.\]
Existence of a neutral element: There exists a number \(1 \in {\mathbb{R}}\) with \(1 \neq 0,\) called “one” such that for all \(a \in {\mathbb{R}}\) we have that \[a \cdot 1 = a.\]
Existence of an inverse element: For every \(a \in {\mathbb{R}}\setminus \{0\}\) there exists a number \(\widetilde{a} \in {\mathbb{R}},\) called multiplicative inverse (element) such that \[a \cdot \widetilde{a} = 1.\]
Axiom of compatibility of addition and multiplication (distributive law): For all \(a,\,b,\,c \in {\mathbb{R}}\) it holds that \[a \cdot (b+c) = (a \cdot b ) + (a \cdot c).\]
Remark 1.1
The neutral elements \(0\) and \(1\) are uniquely determined. Moreover, if \(a \in {\mathbb{R}},\) then its additive inverse is uniquely determined and denoted by \(-a \in {\mathbb{R}}.\) Similarly, \(a \in {\mathbb{R}}\setminus \{ 0 \}\) has a uniquely determined multiplicative inverse which is denoted by \(a^{-1} = \frac{1}{a} \in {\mathbb{R}}.\)
The field axioms are not sufficient in order to characterize the real numbers completely. For example, there are other sets that fulfill all the properties of Axiom 1.1. Typical examples are the sets of rational numbers \({\mathbb{Q}}\) or the set of complex numbers \({\mathbb{C}}\) which will be introduced later in this course. However, the real numbers fulfill additional properties – namely they can be ordered.
Axiom 1.2
There exists a relation “\(<\)” on \({\mathbb{R}},\) called an ordering relation, with the following properties:
Trichotomy: For all \(a,\,b \in {\mathbb{R}}\) exactly one of the following statements is true: \[a<b, \quad a=b, \quad b<a.\]
Transitivity: For all \(a,\,b,\,c \in {\mathbb{R}}\) it holds that \[a < b \text{ and } b < c \quad \Longrightarrow \quad a < c.\]
1st Monotonicity law: For all \(a,\,b,\,c \in {\mathbb{R}}\) it holds that \[a < b \quad \Longrightarrow \quad a + c < b + c.\]
2nd Monotonicity law: For all \(a,\,b,\,c \in {\mathbb{R}}\) it holds that \[a < b \text{ and } 0 < c \quad \Longrightarrow \quad ac < bc.\]
Of course you already know the relation “\(<\)” (“less than”) and we say that \(a\) is less than \(b\) if \(a < b.\) Further, you also know the relations “\(\le\)” (“less or equal than”), “\(>\)” (“greater than”), and “\(\ge\)” (“greater or equal than”). For the latter three we do not need a new axiom since we can reduce these to the relations “\(<\)” and “\(=\)” as follows.
Definition 1.1
Let \(a,\,b \in {\mathbb{R}}.\) Then
\(a \le b \quad :\Longleftrightarrow \quad a<b \text{ or } a=b\);
\(a > b \quad :\Longleftrightarrow \quad b<a\);
\(a \ge b \quad :\Longleftrightarrow \quad b\le a\);
\(a\) is called positive, if \(a > 0\) and negative, if \(a < 0.\)
Unfortunately, Axiom 1.1 and Axiom 1.2 are still not sufficient to characterize the real numbers entirely. For example, the set of rational numbers \({\mathbb{Q}}\) also fulfills all conditions of these two axioms. The question is what precisely makes \({\mathbb{R}}\) different from \({\mathbb{Q}}.\) Are these two sets actually different from each other? To see that \({\mathbb{Q}}\) and \({\mathbb{R}}\) are indeed different sets, we consider the following theorem.
Theorem 1.1
There exists no rational number \(x \in {\mathbb{Q}}\) with the property \(x^2 = 2.\)
Proof. We will prove the statement by deriving a contradiction. Assume that there exists a number \(x \in {\mathbb{Q}}\) with \(x^2 = 2.\) Then by the definition of the set \({\mathbb{Q}},\) there exist \(p,\,q \in {\mathbb{Z}}\) with \(q > 0\) such that \(x = \frac{p}{q}.\) We can assume without loss of generality that \(p\) and \(q\) are not both even numbers, otherwise we could simplify the fraction such that \(p\) or \(q\) is odd. From \(x^2 = \frac{p^2}{q^2} = 2\) we obtain \(p^2 = 2q^2.\) Hence \(p^2\) and also \(p\) are even. Thus there exists an \(m \in {\mathbb{Z}}\) such that \(p = 2m.\) Then we get \[2q^2 = p^2 = 4 m^2,\] and so \(q^2 = 2m^2.\) Hence also \(q^2\) and \(q\) are even. But this is a contradiction because we have assumed that not both \(p\) and \(q\) are even. Hence our assumption \(x \in {\mathbb{Q}}\) is false and there exists no \(x \in {\mathbb{Q}}\) that solves \(x^2 = 2.\)
Of course, we know from school that the solutions of \(x^2 = 2\) are \(-\sqrt{2}\) and \(\sqrt{2},\) but we have just seen that \(\pm\sqrt{2} \notin{\mathbb{Q}}.\) So this means that there are “holes” in the set of rational numbers that are filled by the so-called irrational numbers which appear for instance as solutions of quadratic equations. To fix this problem we will introduce another axiom. For that we need a couple of definitions and preliminary considerations.
Definition 1.2 • Boundedness from above, upper bound
Let \(M \subseteq {\mathbb{R}}.\) The set \(M\) is called bounded from above, if there exists an \(s \in {\mathbb{R}}\) such that \(x \le s\) is satisfied for all \(x \in M.\) In this case, the number \(s \in {\mathbb{R}}\) is called an upper bound of \(M.\)
Example 1.1
The intervals \([a,b],\) \([a,b),\) \((a,b],\) \((a,b),\) \((-\infty,b],\) and \((-\infty,b)\) are all bounded from above and an upper bound is \(b\) in all cases, but also any number \(c \ge b\) is an upper bound.
The interval \([a,\infty)\) is not bounded from above. Assume that \(s\) is an upper bound of \([a,\infty).\) Then \(a \le s\) and hence, \(s \in [a,\infty).\) Since \(s < s+1,\) also \(s+1 \in [a,\infty).\) But then, \(s\) cannot be an upper bound, a contradiction to our assumption.
As seen in Example 1.1, upper bounds are not unique. This motivates the following definition.
Definition 1.3 • Supremum, maximum
Let \(M \subseteq {\mathbb{R}}\) and let \(s_0 \in {\mathbb{R}}\) be an upper bound of \(M.\)
The number \(s_0 \in {\mathbb{R}}\) is called supremum (or smallest upper bound) of \(M,\) if for every upper bound \(s \in {\mathbb{R}}\) it holds that \(s_0 \le s.\) We write \(s_0 = \sup M.\)
If \(s_0\) is a supremum of \(M\) and if additionally, \(s_0 \in M,\) then \(s_0\) is called maximum of \(M.\) We write \(s_0 = \max M.\)
Remark 1.2
Suprema and maxima, if they exist, are unique.
Analogous to the definitions above, for a set \(M \subseteq {\mathbb{R}}\) we can introduce the terms boundedness from below, lower bound, infimum (with the notation \(\inf M\)), and minimum (with the notation \(\min M\)).
Definition 1.4 • Boundedness
A set \(M \in {\mathbb{R}}\) is called bounded, if it is bounded from above and bounded from below, i. e., there exists \(s_1,\,s_2 \in {\mathbb{R}}\) such that for all \(x \in M\) it holds that \[s_1 \le x \le s_2.\]
Example 1.2
The empty set \(\emptyset\) is bounded.
The intervals \([a,b],\) \([a,b),\) \((a,b],\) \((a,b)\) are bounded, the intervals \([a,\infty),\) \((a,\infty),\) \((-\infty,b],\) \((-\infty,b)\) are not.
Consider the set \[M := (-\infty,1) = \{ x \in {\mathbb{R}}\; | \; x<1 \}.\] Then \(\sup M = 1\) (and there exists no maximum because \(1 \notin M\)).
Proof. Assume that \(s < 1.\) Then \[s < \frac{s+1}{2} < 1\] and since \(\frac{s+1}{2} \in M,\) \(s\) is not an upper bound of \(M.\) Thus we have shown that \[s < 1 \quad \Longrightarrow \quad s \text{ is not an upper bound of $M$}.\] By contraposition we arrive at \[s \text{ is an upper bound of $M$} \quad \Longrightarrow \quad s \ge 1.\] Thus \(\sup M = 1.\) \(\Box\)Consider the set \(\widetilde{M} := \left\{x \in {\mathbb{R}}\; \middle| \; x^2 < 2\right\}\) is bounded from above, since \(s = \frac{3}{2}\) is a upper bound of \(\widetilde{M},\) because for all \(x \in {\mathbb{R}}\) we have that \[x > \frac{3}{2} \quad \Longrightarrow \quad x^2 > \frac{9}{4} > 2 \quad \Longrightarrow \quad x \notin \widetilde{M}.\] By contraposition we obtain \[x \in \widetilde{M} \quad \Longrightarrow \quad x \le \frac{3}{2}.\]
Does the set \(\widetilde{M}\) have a supremum? This is not clear a priori, but we will reconsider this question later when considering sequences. For the moment we pose another axiom that will give us the desired property.
Axiom 1.3 • Completeness axiom
Every nonempty set \(M \subseteq {\mathbb{R}}\) that is bounded from above has a supremum.
The next property is an important equivalent characterization of the term “supremum”. From now on, \(c>0\) means that \(c \in {\mathbb{R}}\) and \(c>0.\)
Theorem 1.2
Let \(M \subseteq {\mathbb{R}}\) be nonempty and bounded from above and let \(s \in {\mathbb{R}}.\) Then the following statements are equivalent:
It holds that \(s = \sup M.\)
The number \(s\) is an upper bound of \(M\) and for every \(\varepsilon > 0\) there exists an \(x \in M\) with \(s-\varepsilon < x.\) (In this case we even obtain \(s-\varepsilon < x \le s,\) since \(s\) is an upper bound of \(M.\))
Proof. “i \(\Longrightarrow\) ii”: Suppose \(s = \sup M.\) Then \(s\) is an upper bound of \(M.\) Assume that ii is not true, i. e., there exists an \(\varepsilon > 0\) such that \(x \le s - \varepsilon\) for all \(x \in M.\) But this means that \(s- \varepsilon\) is another other bound and since \(s-\varepsilon < s\) this is a contradiction to the property that \(s\) is the supremum of \(M.\)
“ii \(\Longrightarrow\) i”: Assume that ii is satisfied. Then \(s\) is an upper bound of \(M\) and we must show that \(s\) is indeed that smallest upper bound of \(M.\) Let \(\widetilde{s}\) be another upper bound of \(M.\) We show that \(s \le \widetilde{s}.\) Assume that \(\widetilde{s} < s.\) Then \(\varepsilon := s-\widetilde{s} > 0\) and \(\widetilde{s} = s- \varepsilon.\) By assumption ii, there exists an element \(x \in M\) with \(\widetilde{s} = s - \varepsilon < x.\) This is a contradiction to the assumption that \(\widetilde{s}\) is an upper bound of \(M.\) Hence, \(s \le \widetilde{s}.\) Since these arguments hold true for arbitrary upper bounds \(\widetilde{s},\) this shows that \(s\) is indeed the supremum of \(M.\)
Theorem 1.2 illustrates the interpretation of the supremum as the smallest upper bound. Every decrease of a supremum by \(\varepsilon > 0\) results in \(s - \varepsilon\) not being an upper bound of \(M\) anymore.
Corollary 1.3
Let \(M \subseteq {\mathbb{R}}\) be a nonempty subset and let \(s \in {\mathbb{R}}.\) Then the following statements are equivalent:
It holds that \(s = \inf M.\)
The number \(s\) is a lower bound of \(M\) and for every \(\varepsilon > 0\) there exists an \(x \in M\) with \(s+\varepsilon > x.\)
Video 1.1. Summary: axioms of real numbers.
As an application of the completeness axiom we can now actually prove that the set of natural numbers \({\mathbb{N}}\) is unbounded. This seems to be trivial, but the proof is not, since we cannot derive this property solely with the help of the field and ordering axioms. Indeed, we need the completeness axiom here.
Theorem 1.4
The set \({\mathbb{N}}\) of natural numbers is not bounded from above.
Proof. Suppose \({\mathbb{N}}\) is bounded from above. Since \({\mathbb{N}}\) is a nonempty subset of \({\mathbb{R}},\) by Axiom 1.3 there exists a supremum \(s := \sup {\mathbb{N}}\in {\mathbb{R}}.\) Since \(s\) is an upper bound of \({\mathbb{N}}\) it holds that \[\tag{1.1} n \le s \quad \text{for all } n \in {\mathbb{N}}.\] By Theorem 1.2, for \(\varepsilon = 1,\) there exists an \(n_0 \in {\mathbb{N}}\) with \(s-1 = s- \varepsilon < n_0.\) Therefore, \[s < n_0+1 \in {\mathbb{N}}\] which is a contradiction to (1.1). Therefore, our assumption of boundedness of \({\mathbb{N}}\) is false and thus, \({\mathbb{N}}\) is unbounded from above.
Theorem 1.5 • Archimedean axiom
For two numbers \(a,\,b \in {\mathbb{R}}\) with \(a > 0,\) there exists an \(n \in {\mathbb{N}}\) such that \(a\cdot n > b.\)
Proof. Suppose the statement of the theorem is false. Then we have \(a\cdot n \le b\) for all \(n \in {\mathbb{N}}.\) Because of \(a > 0,\) we obtain \[n \le \frac{b}{a} \quad \text{for all $n \in {\mathbb{N}}.$}\] But then \({\mathbb{N}}\) is bounded from above, a contradiction to Theorem 1.4. Consequently, the statement of the theorem is true.
Note that the Archimedean axiom is actually not an axiom (because we have not used it as fundamental building block, but proved it). The term axiom in this case has historical reasons. In fact, there are also some important theorems that are called a “lemma” because their importance became only clear later.
Video 1.2. Implications of completeness.
Finally, by imposing the completeness axiom, we can also show that every nonempty subset of \({\mathbb{R}}\) that is bounded from below has an infimum.
Theorem 1.6
Every nonempty subset \(M \subseteq {\mathbb{R}}\) that is bounded from below has an infimum.
Proof. Exercise!
2 Sequences
One of the most important concepts in analysis are sequences and the related concepts of limits and convergence. As we will see later, these often play an important role in technical considerations in analysis, but also in numerical analysis where one wants to analyse properties of numerical iteration schemes. In fact, our entire calculus is based on these fundamental concepts. Let us begin with the formal definition of a sequence. In these notes we use the convention \({\mathbb{N}}:= \{0,\,1,\,2,\,\ldots\}.\)
Definition 2.1 • Sequence
A sequence of real numbers \({(a_n)}_{n \in {\mathbb{N}}}\) is a mapping \({\mathbb{N}}\to {\mathbb{R}},\) \(n \mapsto a_n.\) The numbers \(a_0,\,a_1,\,a_2,\,\ldots\) are called elements (or sometimes terms or members) of the sequence \({(a_n)}_{n \in {\mathbb{N}}}.\)
Sometimes we will just use the short notation \((a_n)\) for a sequence, if the index set is clear from context. Sometimes we will also use the longer notation \[{(a_n)}_{n \in {\mathbb{N}}} = (a_0,\,a_1,\,a_2,\,\ldots).\] We will also allow sequences with different index sets. For instance, a sequence may start at the index \(k \in {\mathbb{N}}.\) Then we write \[{(a_n)}_{n \ge k} = (a_k,\,a_{k+1},\,a_{k+2},\,\ldots).\] The latter can be transformed into the original notation by writing \({(a_{n+k})}_{n \in {\mathbb{N}}}.\)
Example 2.1
The sequence \({(a_n)}_{n \in {\mathbb{N}}} = (0,\,1,\,4,\,9,\,16,\,25,\,36,\,49,\,\ldots)\) of square numbers is given by the rule \(a_n = n^2.\)
The sequence \({(a_n)}_{n \in {\mathbb{N}}} = (2,\,3,\,5,\,7,\,11,\,13,\,17,\,\ldots)\) of prime numbers is still a sequence but cannot be described by a simple rule.
The recurrence rule \[\begin{aligned} a_0 &= 1, \quad a_1 = 1, \\ a_{n+1} &= a_n + a_{n-1}, \quad n \ge 1 \end{aligned}\] results in the sequence of Fibonacci numbers \[{(a_n)}_{n \in {\mathbb{N}}} = (1,\,1,\,2,\,3,\,5,\,8,\,13,\,21,\,34,\,55,\,\ldots).\] This sequence has been used to describe the development of a rabbit population over time, i. e., \(n\) describes a time unit (say one month) and \(a_n\) is the number of rabbit couples in month \(n.\)
The sequence given by \(a_n = 1 + \frac{1}{10^n}\) is \[{(a_n)}_{n \in {\mathbb{N}}} = (2,\,1.1,\,1.01,\,1.001,\,1.0001,\,\ldots)\] is special in the sense that it seems to converge to the limit 1.
Also the sequence given by the recurrence rule \[\begin{aligned} a_0 &= 2, \\ a_{n+1} &= \frac{1}{2}\left( a_n + \frac{2}{a_n} \right), \quad n \ge 0 \end{aligned}\] seems to have a limit. Computing the first elements of sequence we obtain \[\begin{alignedat}{3} a_0 &= 2, && \\ a_1 &= \frac{3}{2} &&= 1.5000000000000000, \\ a_2 &= \frac{17}{12}&&\approx 1.4166666666666667, \\ a_3 &= \frac{577}{408} &&\approx 1.4142156862156822, \\ a_4 &= \frac{665857}{470832} && \approx 1.4142135623746899. \end{alignedat}\] As one can see, more and more digits do not change anymore, when the index is growing. In fact, this iteration is an example of a numerical scheme (Heron’s method) for computing the square root of a real number. In our case it computes \(\sqrt{2}\) which is the limit of the sequence above.
Finally, let us consider the sequence \({(a_n)}_{n \ge 1}\) with \[a_n = \left(1 + \frac{1}{n}\right)^n, \quad n \ge 1.\] When we compute the elements of the sequence we get \[\begin{aligned} a_1 &= 2, \\ a_2 &= 2.25, \\ a_3 &= 2.37037037\ldots, \\ a_4 &= 2.44140625, \\ a_5 &= 2.48832, \\ &\ldots \\ a_{10} &= 2.59374246\ldots, \\ &\ldots \\ a_{100} &= 2.70481382\ldots, \\ &\ldots \\ a_{1000} &= 2.71692393\ldots. \end{aligned}\] In fact, by looking at the first elements of the sequence it is not obvious at all whether the sequence has a limit. For this sequence, one can show that its limit is Euler’s number \(\mathrm{e}\), but the sequence converges incredibly slowly.
The examples above may have already given you an intuition what it means that a sequence is converging or that a sequence has a limit, but this intuition may still be ambiguous. So in the next section we want to discuss precise mathematical definitions, because we must be sure what each mathematical term means precisely, before we can analyze its properties.
2.1 Convergence of Sequences
Finding a precise definition of the term “convergence” is actually much more difficult than you would expect. Let us first have a look at different oral descriptions for convergence of a sequence \({(a_n)}_{n \in {\mathbb{N}}}\) to a limit \(a \in {\mathbb{R}}\) that you could potentially think of:
“The sequence \({(a_n)}_{n \in {\mathbb{N}}}\) gets closer and closer to \(a.\)” The sequence \[{(a_n)}_{n \in {\mathbb{N}}} = (2,\,1.5,\,1.25,\,1.125,\,1.0625,\,1.03125,\,\ldots)\] gets closer and closer to the value \(a = 0.\) However, the limit of the sequence seems to be the value \(\widetilde{a} = 1.\)
“The sequence \({(a_n)}_{n \in {\mathbb{N}}}\) gets arbitrarily close to the value \(a\) but never reaches it.” Consider the sequence \[{(a_n)}_{n \in {\mathbb{N}}} = (5,\,4,\,3,\,2,\,1,\,1,\,1,\,1,\,1,\,\ldots)\] This sequence convergences to the value \(a=1,\) but this value is also finally attained by the sequence elements.
“The sequence \({(a_n)}_{n \in {\mathbb{N}}}\) gets arbitrarily close to the value \(a.\)” This eliminates the second counter-example. But let us consider the sequence \[{(a_n)}_{n \in {\mathbb{N}}} = (1,\,0.1,\,1,\,0.01,\,1,\,0.001,\,1,\,0.0001,\,\ldots)\] The sequence gets arbitrarily close to the value \(a=0\) but we still do not consider it as the limit of the sequence, since it jumps again to the value \(1.\)
“The sequence \({(a_n)}_{n \in {\mathbb{N}}}\) gets arbitrarily close to the value \(a\) and closer and closer.” This eliminates the third counter-example, but consider the sequence \[{(a_n)}_{n \in {\mathbb{N}}} = (2,\,0.1,\,0.2,\,0.01,\,0.02,\,0.001,\,0.002,\,0.0001,\,\ldots)\] converges to \(a=0,\) but its elements do not get closer and closer to this value.
As you can see, it is very important to think very carefully how to define a mathematical term. Let us proceed with the formal definition of convergence and limit.
Definition 2.2 • Convergence, divergence, limit, null sequence
Let \({(a_n)}_{n \in {\mathbb{N}}}\) be a sequence of real numbers.
The sequence \({(a_n)}_{n \in {\mathbb{N}}}\) is said to be convergent towards \(a \in {\mathbb{R}}\), if for every \(\varepsilon > 0\) there exists an \(N_\varepsilon \in {\mathbb{N}}\) such that for all \(n \in {\mathbb{N}}\) with \(n \ge N_\varepsilon\) it holds that \[|a_n - a| < \varepsilon.\] In this case, \(a\) is called the limit of the sequence \({(a_n)}_{n \in {\mathbb{N}}}.\) We write \[\lim_{n \to \infty} a_n = a \quad \text{or} \quad a_n \to a \text{ for } n \to \infty.\]
The sequence \({(a_n)}_{n \in {\mathbb{N}}}\) is said to be convergent, if there exists an \(a \in {\mathbb{R}}\) such that it is convergent towards \(a.\) Otherwise, it is called divergent.
The sequence \({(a_n)}_{n \in {\mathbb{N}}}\) is said to be properly divergent towards \(\infty\) (resp. \(-\infty\)), if for every \(M > 0\) there exists an \(N_M \in {\mathbb{N}}\) such that \[a_n > M \text{ (resp. $a_n<-M$) for all } n \ge N_M.\] In this case we write \(\lim_{n \to \infty} a_n = \infty\) (resp. \(\lim_{n \to \infty} a_n = -\infty\)).
The sequence \({(a_n)}_{n \in {\mathbb{N}}}\) is called null sequence, if \({(a_n)}_{n \in {\mathbb{N}}}\) converges towards the limit \(a=0.\)
So what does this abstract definition now mean graphically? If \({(a_n)}_{n \in {\mathbb{N}}}\) converges towards \(a \in {\mathbb{R}},\) then for each \(\varepsilon > 0,\) there exists an \(N_\varepsilon \in {\mathbb{N}}\) such that all elements \(a_n\) of the sequence with \(n \ge N_\varepsilon\) are contained in the \(\varepsilon\)-neighborhood of \(a\) \[U_\varepsilon(a) := \{ x \in {\mathbb{R}}\; | \; |x-a| < \varepsilon \} =(a-\varepsilon,a+\varepsilon).\] In other words: No matter how small \(\varepsilon\) is, there will exist an index \(N_\varepsilon\) (which depends on \(\varepsilon\) – the smaller \(\varepsilon,\) the larger \(N_\varepsilon\) usually is) such that \[a_{N_\varepsilon},\,a_{N_\varepsilon+1},\,a_{N_\varepsilon+2},\,\ldots \in U_\varepsilon(a).\] So for any \(\varepsilon > 0,\) all except of finitely many elements of the sequence will be contained in this \(\varepsilon\)-neighborhood.
Video 2.1. Definition of convergence.
Example 2.2
Constant sequences are convergent. Consider the sequence \({(a_n)}_{n \in {\mathbb{N}}} = (a,\,a,\,a,\,a,\,\ldots)\) for fixed \(a \in {\mathbb{R}}.\) Then \(\lim_{n\to\infty} a_n = a\) because \[|a_n - a| = 0 < \varepsilon\] for all \(n \in {\mathbb{N}}\) and \(\varepsilon > 0.\) Hence, for any \(\varepsilon > 0\) we can choose \(N_\varepsilon = 0.\)
A classical example for a non-constant convergent sequence is the sequence \({(a_n)}_{n \ge 1}\) with \(a_n = \frac{1}{n}.\) We have \[\lim_{n \to \infty} \frac{1}{n} = 0,\] so \({(a_n)}_{n \ge 1}\) is even a null sequence.
Proof. Let \(\varepsilon > 0\) be arbitrary. Then by Theorem 1.5 there exists a number \(N_\varepsilon \in {\mathbb{N}}\) such that \(\varepsilon N_\varepsilon > 1.\) Then for all \(n \ge N_\varepsilon\) we obtain \[\left| \frac{1}{n} - 0 \right| = \frac{1}{n} \le \frac{1}{N_\varepsilon} < \varepsilon.\] Since \(\varepsilon > 0\) is assumed to be arbitrary it follows that 0 is the limit of the sequence \(\big(\frac{1}{n}\big)_{n \ge 1}.\) \(\Box\)
For a given \(\varepsilon > 0\) we can easily state an appropriate \(N_\varepsilon\) – we only need to have \(\varepsilon N_\varepsilon >1.\) For instance, if \(\varepsilon = 0.1\) we could choose \(N_\varepsilon = 11.\)An example for a divergent sequence is \({(a_n)}_{n \in {\mathbb{N}}}\) with \(a_n = (-1)^n,\) that is, \[{(a_n)}_{n \in {\mathbb{N}}} = (1,\,-1,\,1,\,-1,\,1,\,-1,\,\ldots).\] We show that this sequence is divergent.
Proof. Assume that there exists an \(a \in {\mathbb{R}}\) such that \(\lim_{n \to \infty} a_n = a.\) Then for \(\varepsilon = 1>0\) there exists an \(N_1 \in {\mathbb{N}}\) such that \(|a_n - a| < 1\) for all \(n \ge N_1.\) By using the triangle inequality we obtain \[\begin{aligned} 2 = |a_{n+1} - a_n| &= |(a_{n+1} - a) + (a - a_n)| \\ &\le |a_{n+1} - a| + |a - a_n| < 1+1 = 2. \end{aligned}\] Thus we obtain \(2 < 2\) which is a contradiction. Hence our original assumption of convergence towards \(a\) is false and therefore, the sequence must be divergent. \(\Box\)The sequence \({(a_n)}_{n \in {\mathbb{N}}}\) with \(a_n = n\) is properly divergent towards \(\infty.\)
Proof. Let \(M > 0\) be arbitrary. Set \(N_M := \lceil M \rceil+1,\) where \(\lceil M \rceil\) denotes the smallest integer that is larger or equal to \(M\) (which exists due to Theorem 1.4). Then we have \[a_n \ge \lceil M \rceil+1 > M \quad \text{for all }n \ge N_M,\] which proves the assertion. \(\Box\)
Video 2.2. Examples for convergence and divergence.
Up to now we have only talked about the limit, but is not clear a priori that the limit is uniquely determined. Hence, in the next theorem we will prove that the limit is indeed unique.
Theorem 2.1
A convergent sequence \({(a_n)}_{n \in {\mathbb{N}}}\) of real numbers has exactly one limit.
Proof. Let \(a\) and \(b\) be limits of the sequence \({(a_n)}_{n \in {\mathbb{N}}}\) and assume that \(a\neq b.\) Then for \(\varepsilon := \frac{|a-b|}{2} > 0\) there exist \(\widehat{N}_\varepsilon,\,\widetilde{N}_\varepsilon \in {\mathbb{N}}\) such that \[\begin{aligned} |a_n - a| &< \varepsilon \quad \text{for all } n \ge \widehat{N}_\varepsilon, \\ |a_n - b| &< \varepsilon \quad \text{for all } n \ge \widetilde{N}_\varepsilon. \end{aligned}\] As in Example 2.2 ii we will try to find a contradication. Now set \(N_\varepsilon := \max\big\{\widehat{N}_\varepsilon ,\,\widetilde{N}_\varepsilon \big\}.\) Then for \(n \ge N_\varepsilon\) and by using the triangle inequality we get \[\begin{aligned} |a-b| = |a - a_n + a_n -b| \le |a-a_n| + |a_n - b| < \varepsilon + \varepsilon = 2\varepsilon = |a-b|. \end{aligned}\] Hence \(|a-b| < |a-b|\) which is a contradiction and so we have \(a = b.\)
Video 2.3. Uniqueness of limits.
Remark 2.1
From every convergent sequence we can easily obtain a null sequence. It follows directly from Definition 2.2 that \({(a_n)}_{n \in {\mathbb{N}}}\) convergences towards \(a \in {\mathbb{R}},\) if and only if \({(a_n-a)}_{n \in {\mathbb{N}}}\) is a null sequence.
For the analysis of convergence of sequences we will now consider a necessary criterion.
Definition 2.3 • Bounded sequence
A sequence \({(a_n)}_{n \in {\mathbb{N}}}\) is called bounded, if the set \(\{ a_n \;|\;n\in{\mathbb{N}}\}\) is bounded, that is, there exists a real number \(M>0\) such that \[|a_n| \le M \quad \text{for all } n\in{\mathbb{N}}.\]
Theorem 2.2
Every convergent sequence of real numbers is bounded.
Proof. Let \({(a_n)}_{n \in {\mathbb{N}}}\) be a convergent sequence with limit \(a \in {\mathbb{R}}.\) Then for \(\varepsilon:=1>0\) there exist an \(N_1 \in {\mathbb{N}}\) such that \(|a_n - a| < 1\) for all \(n \ge N_1.\) Then with the help of the triangle inequality we obtain \[|a_n| = |a + a_n -a | \le |a| + |a_n-a| < |a| + 1\] for all \(n \ge N_1.\) If we set \[M:= \max \{ |a_0|,\,|a_1|,\,\ldots,\,|a_{N_1-1}|, |a|+1 \},\] we obtain a bound for \({(a_n)}_{n \in {\mathbb{N}}},\) i. e., \(|a_n| \le M\) for all \(n \in {\mathbb{N}}\) and so the sequence is bounded.
Video 2.4. Boundedness of convergent sequences.
Example 2.3
Theorem 2.2 now delivers a simple cirterion to decide when a sequence is not convergent, namely if it is not bounded. Let us have a look at two examples.
The sequence \({(n)}_{n \in {\mathbb{N}}}\) is divergent because it is not bounded since \({\mathbb{N}}\) is not bounded.
Let \(x \in {\mathbb{R}}\) and consider the sequence \(\big(x^n\big)_{n \in {\mathbb{N}}}\) and distinguish three cases:
Case 1: \(|x|>1.\) Let \(M\) be arbitrary and \(y:=|x|-1.\) Then we have \(y > 0.\) Then for \(n \ge 1\) we obtain \[\begin{aligned} \left|x^n\right| = |x|^n = (1+y)^n = \sum_{k=0}^n\begin{pmatrix} n \\ k \end{pmatrix} 1^{n-k} y^k \ge \begin{pmatrix} n \\ 1 \end{pmatrix} 1^{n-1} y^1 = ny \end{aligned}\] Here we have used the binomial theorem for expressing arbitrary powers of \((1+y)\) by binomial coefficients. Furthermore, for the inequality we have used the fact that all summands in the sum are positive. By Theorem 1.5 there exists an \(N \in {\mathbb{N}}\) with \(Ny > M\) and hence, \[\big|x^N\big| \ge Ny > M.\] Therefore, since \(M\) is arbitrary, the sequence \(\big(x^n\big)_{n \in {\mathbb{N}}}\) is not bounded and hence, it cannot be convergent.
Case 2: \(|x|=1.\) If \(x=1,\) then \(\big(x^n\big)_{n \in {\mathbb{N}}}\) is constant and converging. If \(x=-1,\) we obtain the divergent sequence \(\big(x^n\big)_{n \in {\mathbb{N}}} = \big((-1)^n\big)_{n \in {\mathbb{N}}}.\)
Case 3: \(|x|<1.\) In this case, \(\frac{1}{|x|} > 1\) and thus \(y := \frac{1}{|x|} -1 >0\) and similarly to Case 1 we get \[\frac{1}{|x|^n} = \left( \frac{1}{|x|} \right)^n = (1+y)^n \ge ny.\] Then we get \(\big|x^n\big| = |x|^n \le \frac{1}{ny}.\) Hence \(\big|x^n\big| \le \frac{1}{y}\) for all \(n \ge 1\) and the sequence \(\big(x^n\big)_{n \in {\mathbb{N}}}\) is bounded. However, we still do not know whether it is also convergent. So in the next step we will actually show that \(\lim_{n \to \infty} x^n = 0.\)
Proof. Let \(\varepsilon > 0.\) Since \(\big(\frac{1}{n}\big)_{n \ge 1}\) is a null sequence, for \(\widetilde{\varepsilon} := \varepsilon y\) there exists an \(N_{\widetilde{\varepsilon}} \in {\mathbb{N}}\) such that \[\left| \frac{1}{n} - 0 \right| = \frac{1}{n} < \varepsilon y\] for all \(n \ge N_{\widetilde{\varepsilon}}.\) Then we obtain \[|x|^n \le \frac{1}{y} \cdot \frac{1}{n} < \frac{1}{y}\varepsilon y = \varepsilon\] for all \(n \ge N_{\widetilde{\varepsilon}}.\) Since \(\varepsilon > 0\) is arbitrary, the claim follows. \(\Box\)