2.1: Convergence
( \newcommand{\kernel}{\mathrm{null}\,}\)
Let {an} be a sequence of real numbers. We say that the sequence {an} converges to a∈R if, for any ε>0, there exists a positive integer N such that for any n∈N with n≥N, one has
|an−a|<ε( or equivalently ,a−ε<an<a+ε).
In this case, we call a the limit of the sequence (see Theorem 2.1.3 below) and write limn→∞an=a. If the sequence {an} does not converge, we call the sequence divergent.
It follows directly from the definition, using the Archimedean property, that a sequence {an} converges to a if and only if for any ε>0, there exists a real number N such that for any n∈N with n>N, one has
|an−a|<ε.
Let an=1n for n∈N. We claim that limn→∞an=0. We verify it using the definition.
Solution
Let ε>0. Choose an integer N>1/ε. (Note that such an integer N exists due to the Archimidean Property.) Then, if n≥N, we get
|an−0|=|1n|=1n≤1N<11/ε=ε.
We now generalize the previous example as follows. Let α>0 and consider the sequence given by
an=1nα for n∈N.
Solution
Let ε>0. Choose an integer N such that N>(1ε)1/α. For every n≥N, one has n>(1ε)1/α and, hence, nα>1ε. This implies
|1nα−0|=1nα<11/ε=ε.
We conclude that limn→∞an=0.
Consider the sequence {an} where
an=3n2+42n2+n+5
We will prove directly from the definition that this sequence converges to a=32.
Solution
Let ε>0. We first search for a suitable N. To that end, we simplify and estimate the expression |an−a|. Notice that
|an−32|=|3n2+42n2+n+5−32|=|2(3n2+4)−3(2n2+n+5)2(2n2+n+5)|=|−7−3n2(2n2+n+5)|=3n+72(2n2+n+5)<10n4n2=104n
To guarantee that the last expression is less than ε, it will suffice to choose N>104ε. Indeed, if n≥N, we get
|an−a|≤104n≤104N<104104ε=ε.
Let {an} be given by
an=4n2−13n2−n.
We claim limn→∞an=43.
Solution
Let ε>0. We search for a suitable N. First notice that
|4n2−13n2−n−43|=|12n2−3−12n2+4n3(3n2−n)|=|4n−33(3n2−n)|
Since n≥1, we have n2≥n and 4n>3. Therefore we have
|4n2−13n2−n−43|=4n−33(3n2−n)≤4n−33(3n2−n2)<4n6n2=46n.
Thus, if N>46ε, we have, for n≥N
|4n2−13n2−n−43|≤46n≤46N<ε.
Consider the sequence given by
an=n2+54n2+n.
We prove directly from the definition that {an} converges to 14.
Solution
Let ε>0. Now,
|n2+54n2+n−14|=|4n2+20−4n2−n4(4n2+n)|=|20−n|4(4n2+n).
If n≥20, then |20−n|=n−20. Therfore, for such n we have
|n2+54n2+n−14|=n−204(4n2+n)≤n16n2=116n.
Choose N>max{116ε,20}. Then, for n≥N we get
|n2+54n2+n−14|≤116n≤116N<ε.
The following result is quite useful in proving certain inequalities between numbers.
Let ℓ≥0, If ℓ<ε for all ε>0, then ℓ=0.
- Proof
-
This is easily proved by contraposition. If ℓ>0, then there is a positive number, for example ε=ℓ/2, such that ε<ℓ. ◻
A convergent sequence {an} has at most one limit
- Proof
-
Suppose {an} converges to a and b. Then given ε>0, there exist positive integers N1 and N2 such that
|an−a|<ε/2 for all n≥N1
and
|an−b|<ε/2 for all n≥N2.
Let N=max{N1,N2}. Then
|a−b|≤|a−aN|+|aN−b|<ε/2+ε/2=ε.
Since ε>0 is arbitrary, by Lemma 2.1.2, |a−b|=0 and, hence, a=b. ◻
The following lemma is simple generalization of (2.1.2).
Given real numbers a,b, then a≤b if and only if a<b+ε for all ε>0.
- Proof
-
Suppose a<b+ε for all ε>0. And suppose, by way of contradiction, that a>b. then set ε0=a−b. Then ε0>0. By assumption, we should have a<b+ε0=b+a−b=a, which is a contradiction. It follows that a≤b.
The other direction follows immediately from the order axioms. ◻
The following comparison theorem shows that (non-strict) inequalities are preserved "in the limit".
Suppose {an} and {bn} converge to a and b, respectively, and an≤bn for all n∈N. Then a≤b.
- Proof
-
For any ε>0, there exist N1,N2∈N such that
a−ε2<an<a+ε2, for n≥N1,
b−ε2<bn<b+ε2, for n≥N2.
Choose N>max{N1,N2}. Then
a−ε2<aN≤bN<b+ε2.
Thus, a<b+ε for any ε>0. Using Lemma 2.1.4 we conclude a≤b. ◻
Suppose the sequences {an}, {bn}, and {cn} satisfy
an≤bn≤cn for all n∈N,
and limn→∞an=limn→∞cn=ℓ. Then limn→∞bn=ℓ.
- Proof
-
Fix any ε>0. Since limn→∞an=ℓ, there exists N1∈N such that
ℓ−ε<an<ℓ+ε
for all n≥N1. Similarly, since limn→∞cn=ℓ, there exists N2∈N such that
ℓ−ε<cn<ℓ+ε
for all n≥N2. Let N=max{N1,N2}. Then, for n≥N, we have
ℓ−ε<an≤bn≤cn<ℓ+ε,
which implies |bn−ℓ|<ε. Therefore, limn→∞bn=ℓ. ◻
A sequence {an} is bounded above if the set {an:n∈N} is bounded above. Similarly, the sequence \left\{a_{n}: n \in \mathbb{N}\right\} is bounded below if the set {an:n∈N} is bounded below. We say that the sequence {an} is bounded if the set {an:n∈N} is bounded, that is, if it is both bounded above and bounded below.
It follows from the observation after Definition 1.5.1 that the sequence {an} is bounded if and only if there is M∈R such that |an|≤M for all n∈N.
A convergent sequence is bounded.
- Proof
-
Suppose the sequence {an} converges to a. Then, for ε=1, there exists N∈N such that
|an−a|<1 for all n≥N.
Since |an|−|a|≤‖, this implies \left|a_{n}\right|<1+|a| for all n \geq N. Set
M=\max \left\{\left|a_{1}\right|, \ldots,\left|a_{N-1}\right|,|a|+1\right\}
Then \left|a_{n}\right| \leq M for all n \in \mathbb{N}. Therefore, \left\{a_{n}\right\} is bounded. \square
Let \left\{a_{n}\right\}_{n=1}^{\infty} be a sequence of real numbers. The sequence \left\{b_{n}\right\}_{n=1}^{\infty} is called a subsequence of \left\{a_{n}\right\}_{n=1}^{\infty} if there exists a sequence of increasing positive integers
n_{1}<n_{2}<n_{3}<\cdots,
such that b_{k}=a_{n_{k}} for each k \in \mathbb{N}.
Consider the sequence a_{n}=(-1)^{n} for n \in \mathbb{N}.
Solution
Then \left\{a_{2 k}\right\} is a subsequence of \left\{a_{n}\right\} and a_{2 k}=1 for all k (here n_{k}=2 k for all k). Similarly, \left\{a_{2 k+1}\right\} is also a subsequence of \left\{a_{n}\right\} and a_{2 k+1}=-1 for all k (here n_{k}=2 k +1 for all k).
Let \left\{n_{k}\right\}_{k} be a sequence of positive integers with
n_{1}<n_{2}<n_{3}<\cdots
Then n_{k} \geq k for all k \in \mathbb{N}.
- Proof
-
We use mathematical induction. When k=1, it is clear that n_{1} \geq 1 since n_{1} is a positive integer. Assume n_{k} \geq k for some k. Now n_{k+1}>n_{k} and, since n_{k} and n_{k+1} are integers, this implies, n_{k+1} \geq n_{k}+1. Therefore, n_{k+1} \geq k+1 by the inductive hyposthesis. The conclusion now follows by the principle of mathematical induction. \square
If a sequence \left\{a_{n}\right\} converges to a, then any subsequence \left\{a_{n_{k}}\right\} of \left\{a_{n}\right\} also converges to a.
- Proof
-
Suppose \left\{a_{n}\right\} converges to a and let \varepsilon>0 be given. Then there exists N such that
\left|a_{n}-a\right|<\varepsilon \text { for all } n \geq N.
For any k \geq N, since n_{k} \geq k, we also have
\left|a_{n_{k}}-a\right|<\varepsilon.
Thus, \left\{a_{n_{k}}\right\} converges to a as k \rightarrow \infty. \square
Let a_{n}=(-1)^{n} for n \in \mathbb{N}.
Solution
Then the sequence \left\{a_{n}\right\} is divergent. Indeed suppose by contradiction that
\lim _{n \rightarrow \infty} a_{n}=\ell.
Then every subsequence of \left\{a_{n}\right\} converges to a number \ell \in \mathbb{R}. From the previous theorem, it follows, in particular, that
\ell=\lim _{k \rightarrow \infty} a_{2 k}=1 \text { and } \ell=\lim _{k \rightarrow \infty} a_{2 k+1}=-1.
This contradiction shows that the sequence is divergent.
Since the sequence \left\{a_{n}\right\} is bounded but not convergent, this example illustrates the fact that the converse of theorem 2.1.7 is not true.
Given a positive integer k_{0}, it will be convenient to also talk about the sequence \left\{a_{n}\right\}_{n \geq k_{0}}, that is, a function defined only for the integers greater than or equal to k_{0}. For simplicity of notation, we may also denote this sequence by \left\{a_{n}\right\} whenever the integer k_{0} is clear form the context. For instance, we talk of the sequence \left\{a_{n}\right\} given by
a_{n}=\frac{n+1}{(n-1)(n-2)}.
although a_{1} and a_{2} are not defined. In all cases, the sequence must be defined from some integer onwards.
Exercise \PageIndex{1}
Prove the following directly from the definition of limit.
- \lim _{n \rightarrow \infty} \frac{2 n^{2}+2}{3 n^{3}+1}=0.
- \lim _{n \rightarrow \infty} \frac{n^{2}+1}{5 n^{2}+n+1}=\frac{1}{5}.
- \lim _{n \rightarrow \infty} \frac{2 n^{3}+1}{4 n^{3}-n}=\frac{1}{2}.
- \lim _{n \rightarrow \infty} \frac{3 n^{2}+5}{6 n^{2}+n}=\frac{1}{2}.
- \lim _{n \rightarrow \infty} \frac{4 n^{2}-1}{n^{2}-n}=4.
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{2}
Prove that if \left\{a_{n}\right\} is a convergent sequence, then \left\{|a_{n}|\right\} is a convergent sequence. Is the converse true?
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{3}
Let \left\{a_{n}\right\} be a sequence. Prove that if the \left\{|a_{n}|\right\} converges to 0, then \left\{a_{n}\right\} also converges to 0.
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{4}
Prove that \lim _{n \rightarrow \infty} \frac{\sin n}{n}=0.
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{5}
Let \left\{x_{n}\right\} be a bounded sequence and let \left\{y_{n}\right\} be a sequence that converges to 0. Prove that the sequence \left\{x_{n} y_{n}\right\} converges to 0.
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{6}
Prove that the following limits are 0. (Hint: use Theorem 2.1.6.)
- \lim _{n \rightarrow \infty} \frac{n+\cos \left(n^{2}-3\right)}{2 n^{2}+1}.
- \lim _{n \rightarrow \infty} \frac{3^{n}}{n !}.
- \lim _{n \rightarrow \infty} \frac{n !}{n^{n}}.
- \lim _{n \rightarrow \infty} \frac{n^{2}}{3^{n}}. (Hint: see Exercise 1.3.4(c)).
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{7}
Prove that if \lim _{n \rightarrow \infty} a_{n}=\ell>0, then there exists N \in \mathbb{N} such that a_{n}>0 for all n \geq N.
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{8}
Prove that if \lim _{n \rightarrow \infty} a_{n}=\ell \neq 0, then \lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=1. Is the conclusion still true if \ell =0?
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{9}
Let \left\{a_{n}\right\} be a sequence of real numbers such that \lim _{n \rightarrow \infty} a_{n}=3. Use Definition 2.1.1 to prove the following
- \lim _{n \rightarrow \infty} 3 a_{n}-7=2;
- \lim _{n \rightarrow \infty} \frac{a_{n}+1}{a_{n}}=\frac{4}{3}; (Hint: prove first that there is N such that a_{n}>1 for n \geq N.)
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{10}
Let a_{n} \geq 0 for all n \in \mathbb{N}. Prove that if \lim _{n \rightarrow \infty} a_{n}=\ell, then \lim _{n \rightarrow \infty} \sqrt{a_{n}}=\sqrt{\ell}.
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{11}
Prove that the sequence \left\{a_{n}\right\} with a_{n}=\sin (n \pi / 2) is divergent.
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{12}
Consider a sequence \left\{a_{n}\right\}.
- Prove that \lim _{n \rightarrow \infty} a_{n}=\ell if and only if \lim _{k \rightarrow \infty} a_{2 k}=\ell and \lim _{k \rightarrow \infty} a_{2 k+1}=\ell.
- Prove that \lim _{n \rightarrow \infty} a_{n}=\ell if and only if \lim _{k \rightarrow \infty} a_{3 k}=\ell, \lim _{k \rightarrow \infty} a_{3 k+1}=\ell, and \lim _{k \rightarrow \infty} a_{3 k+2}=\ell.
- Answer
-
Add texts here. Do not delete this text first.
Exercise \PageIndex{13}
Given a sequence \left\{a_{n}\right\}, define a new sequence \left\{b_{n}\right\} by
b_{n}=\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}.
- Prove that if \lim _{n \rightarrow \infty} a_{n}=\ell, then \lim _{n \rightarrow \infty} b_{n}=\ell
- Find a counterexample to show that the converse does not hold in general.
- Answer
-
Add texts here. Do not delete this text first.