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2.1: Convergence

( \newcommand{\kernel}{\mathrm{null}\,}\)

Definition 2.1.1

Let {an} be a sequence of real numbers. We say that the sequence {an} converges to aR if, for any ε>0, there exists a positive integer N such that for any nN with nN, one has

|ana|<ε( or equivalently ,aε<an<a+ε).

In this case, we call a the limit of the sequence (see Theorem 2.1.3 below) and write limnan=a. If the sequence {an} does not converge, we call the sequence divergent.

Remark 2.1.1

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 nN with n>N, one has

|ana|<ε.

Example 2.1.1

Let an=1n for nN. We claim that limnan=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 nN, we get

|an0|=|1n|=1n1N<11/ε=ε.

Example 2.1.2

We now generalize the previous example as follows. Let α>0 and consider the sequence given by

an=1nα for nN.

Solution

Let ε>0. Choose an integer N such that N>(1ε)1/α. For every nN, one has n>(1ε)1/α and, hence, nα>1ε. This implies

|1nα0|=1nα<11/ε=ε.

We conclude that limnan=0.

Example 2.1.3

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 |ana|. Notice that

|an32|=|3n2+42n2+n+532|=|2(3n2+4)3(2n2+n+5)2(2n2+n+5)|=|73n2(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 nN, we get

|ana|104n104N<104104ε=ε.

Example 2.1.4

Let {an} be given by

an=4n213n2n.

We claim limnan=43.

Solution

Let ε>0. We search for a suitable N. First notice that

|4n213n2n43|=|12n2312n2+4n3(3n2n)|=|4n33(3n2n)|

Since n1, we have n2n and 4n>3. Therefore we have

|4n213n2n43|=4n33(3n2n)4n33(3n2n2)<4n6n2=46n.

Thus, if N>46ε, we have, for nN

|4n213n2n43|46n46N<ε.

Example 2.1.5

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+n14|=|4n2+204n2n4(4n2+n)|=|20n|4(4n2+n).

If n20, then |20n|=n20. Therfore, for such n we have

|n2+54n2+n14|=n204(4n2+n)n16n2=116n.

Choose N>max{116ε,20}. Then, for nN we get

|n2+54n2+n14|116n116N<ε.

The following result is quite useful in proving certain inequalities between numbers.

Lemma 2.1.2

Let 0, If <ε for all ε>0, then =0.

Proof

This is easily proved by contraposition. If >0, then there is a positive number, for example ε=/2, such that ε<.

Theorem 2.1.3

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

|ana|<ε/2 for all nN1

and

|anb|<ε/2 for all nN2.

Let N=max{N1,N2}. Then

|ab||aaN|+|aNb|<ε/2+ε/2=ε.

Since ε>0 is arbitrary, by Lemma 2.1.2, |ab|=0 and, hence, a=b.

The following lemma is simple generalization of (2.1.2).

Lemma 2.1.4

Given real numbers a,b, then ab 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=ab. Then ε0>0. By assumption, we should have a<b+ε0=b+ab=a, which is a contradiction. It follows that ab.

The other direction follows immediately from the order axioms.

The following comparison theorem shows that (non-strict) inequalities are preserved "in the limit".

Theorem 2.1.5 - Comparison Theorem.

Suppose {an} and {bn} converge to a and b, respectively, and anbn for all nN. Then ab.

Proof

For any ε>0, there exist N1,N2N such that

aε2<an<a+ε2, for nN1,

bε2<bn<b+ε2, for nN2.

Choose N>max{N1,N2}. Then

aε2<aNbN<b+ε2.

Thus, a<b+ε for any ε>0. Using Lemma 2.1.4 we conclude ab.

Theorem 2.1.6 - The Squeeze Theorem.

Suppose the sequences {an}, {bn}, and {cn} satisfy

anbncn for all nN,

and limnan=limncn=. Then limnbn=.

Proof

Fix any ε>0. Since limnan=, there exists N1N such that

ε<an<+ε

for all nN1. Similarly, since limncn=, there exists N2N such that

ε<cn<+ε

for all nN2. Let N=max{N1,N2}. Then, for nN, we have

ε<anbncn<+ε,

which implies |bn|<ε. Therefore, limnbn=.

Definition 2.1.2

A sequence {an} is bounded above if the set {an:nN} is bounded above. Similarly, the sequence \left\{a_{n}: n \in \mathbb{N}\right\} is bounded below if the set {an:nN} is bounded below. We say that the sequence {an} is bounded if the set {an:nN} 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 MR such that |an|M for all nN.

Theorem 2.1.7

A convergent sequence is bounded.

Proof

Suppose the sequence {an} converges to a. Then, for ε=1, there exists NN such that

|ana|<1 for all nN.

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

Definition \PageIndex{3}

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}.

Example \PageIndex{6}

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).

Lemma \PageIndex{8}

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

Theorem \PageIndex{9}

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

Example \PageIndex{1}

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.

Remark \PageIndex{10}

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.

  1. \lim _{n \rightarrow \infty} \frac{2 n^{2}+2}{3 n^{3}+1}=0.
  2. \lim _{n \rightarrow \infty} \frac{n^{2}+1}{5 n^{2}+n+1}=\frac{1}{5}.
  3. \lim _{n \rightarrow \infty} \frac{2 n^{3}+1}{4 n^{3}-n}=\frac{1}{2}.
  4. \lim _{n \rightarrow \infty} \frac{3 n^{2}+5}{6 n^{2}+n}=\frac{1}{2}.
  5. \lim _{n \rightarrow \infty} \frac{4 n^{2}-1}{n^{2}-n}=4.
Answer

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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

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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

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Exercise \PageIndex{4}

Prove that \lim _{n \rightarrow \infty} \frac{\sin n}{n}=0.

Answer

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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

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Exercise \PageIndex{6}

Prove that the following limits are 0. (Hint: use Theorem 2.1.6.)

  1. \lim _{n \rightarrow \infty} \frac{n+\cos \left(n^{2}-3\right)}{2 n^{2}+1}.
  2. \lim _{n \rightarrow \infty} \frac{3^{n}}{n !}.
  3. \lim _{n \rightarrow \infty} \frac{n !}{n^{n}}.
  4. \lim _{n \rightarrow \infty} \frac{n^{2}}{3^{n}}. (Hint: see Exercise 1.3.4(c)).
Answer

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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

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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

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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

  1. \lim _{n \rightarrow \infty} 3 a_{n}-7=2;
  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

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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

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Exercise \PageIndex{11}

Prove that the sequence \left\{a_{n}\right\} with a_{n}=\sin (n \pi / 2) is divergent.

Answer

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Exercise \PageIndex{12}

Consider a sequence \left\{a_{n}\right\}.

  1. 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.
  2. 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

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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}.

  1. Prove that if \lim _{n \rightarrow \infty} a_{n}=\ell, then \lim _{n \rightarrow \infty} b_{n}=\ell
  2. Find a counterexample to show that the converse does not hold in general.
Answer

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This page titled 2.1: Convergence is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Lafferriere, Lafferriere, and Nguyen (PDXOpen: Open Educational Resources) .

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