Sequences | An Introduction to Real Analysis

In the tool box used to build analysis, if the Completeness property of the real numbers is the hammer then sequences are the nails. Almost everything that can be said in analysis can be, and is, done using sequences. For this reason, the study of sequences will occupy us for the next foreseeable future.

Limits of Sequences

A sequence of real numbers is a function

X : N \to R

. Informally, the sequence

X

can be written as an infinite list of real numbers as

X = (x_{1}, x_{2}, x_{3}, \dots)

, where

x_{n} = X (n)

. Other notations for sequences are

(x_{n})

{x_{n}}_{n = 1}^{\infty}

; we will use

(x_{n})

. Some sequences can be written explicitly with a formula such as

x_{n} = \frac{1}{n}

x_{n} = \frac{1}{2^{n}}

, or

x_{n} = (- 1)^{n} \cos (n^{2} + 1),

or we could be given the first few terms of the sequence, such as

X = (3, 3.1, 3.14, 3.141, 3.1415, \dots) .

Some sequences may be given recursively. For example,

x_{1} = 1, x_{n + 1} = \frac{x_{n}}{n + 1}, n \geq 1.

Using the definition of

x_{n + 1}

and the initial value

x_{1}

we can in principle find all the terms:

x_{2} = \frac{1}{2}, x_{3} = \frac{1 / 2}{3}, x_{4} = \frac{1 / 6}{4}, \dots

A famous sequence given recursively is the Fibonacci sequence which is defined as

x_{1} = 1

x_{2} = 1

, and

x_{n + 1} = x_{n - 1} + x_{n}, n \geq 2.

Then

x_{3} = 2, x_{4} = 3, x_{5} = 5, \dots

The range of a sequence

(x_{n})

is the set

{x_{n} | n \in N},

that is, the usual range of a function. However, the range of a sequence is not the actual sequence (the range is a set and a sequence is a function). For example, if

X = (1, 2, 3, 1, 2, 3, \dots)

then the range of

X

{1, 2, 3}

. If

x_{n} = \sin (\frac{n π}{2})

then the range of

(x_{n})

{1, 0, - 1}

. Many concepts in analysis can be described using the long-term or limiting behavior of sequences. In calculus, you undoubtedly developed techniques to compute the limit of basic sequences (and hence show convergence) but you might have omitted the rigorous definition of the convergence of a sequence. Perhaps you were told that a given sequence

(x_{n})

converges to

L

if as

n \to \infty

the values

x_{n}

get closer to

L

. Although this is intuitively sound, we need a more precise way to describe the meaning of the convergence of a sequence. Before we give the precise definition, we will consider an example.

Consider the sequence

(x_{n})

whose

n

th term is given by

x_{n} = \frac{3 n + 2}{n + 1}

. The values of

(x_{n})

for several values of

n

are displayed in Table 3.1.

$n$	$x_{n}$
1	2.50000000
2	2.66666667
3	2.75000000
4	2.80000000
5	2.83333333
50	2.98039216
101	2.99019608
10,000	2.99990001
1,000,000	2.99999900
2,000,000	2.99999950

Values of the sequence

The above data suggests that the values of the sequence

become closer and closer to the number

. For example, suppose that

and consider the

-neighborhood of

, that is, the interval

. Not all the terms of the sequence

are in the

-neighborhood, however, it seems that all the terms of the sequence from

and onward are inside the

-neighborhood. In other words, IF

then

, or equivalently

. Suppose now that

and thus the new

-neighborhood is

. Then it is no longer true that

for all

. However, it seems that all the terms of the sequence from

and onward are inside the smaller

-neighborhood, in other words,

for all

. We can extrapolate these findings and make the following hypothesis: For any given

there exists a natural number

such that if

then

The above example and our analysis motivates the following definition.

The sequence

is said to converge if there exists a number

such that for any given

there exists

such that

for all

. In this case, we say that

has limit

and we write

is not convergent then we say that it is divergent.

Hence,

converges to

if for any given

(no matter how small), there exists a point in the sequence

such that

, that is,

for all

. We will sometimes write

simply as

Using the definition of the limit of a sequence, prove that

Let

be arbitrary but fixed. By the Archimedean property of

, there exists

such that

. Then, if

then

. Therefore, if

then

This proves, by definition, that

Using the definition of the limit of a sequence, prove that

Given an arbitrary

, we want to prove that there exists

such that

Start by analyzing

Now, the condition that

it is equivalent to

Now let's write the formal proof. Let

be arbitrary and let

be such that

. Then,

. Now if

then

and thus if

then

By definition, this proves that

Using the definition of the limit of a sequence, prove that

Let

. We want to show that for any given

, there exists

such that if

then

Start by analyzing

In this case, it is difficult to explicitly isolate for

in terms of

. Instead we take a different approach; we find an upper bound for

Hence, if

then also

by the transitivity property of inequalities. The inequality

holds true if and only if

. Now that we have done a detailed preliminary analysis, we can proceed with the proof. Suppose that

is given and let

be such that

. Then

, and thus

. Then, if

then

and therefore

This proves that

Prove that for any irrational number

there exists a sequence of rational numbers

converging to

Let

be any sequence of positive numbers converging to zero, for example,

. Now since

for each

, then by the Density theorem there exists a rational number

such that

. In other words,

. Now let

be arbitrary. Since

converges to zero, there exists

such that

for all

, or since

, then

for all

. Therefore, if

then

. Thus, for arbitrary

there exists

such that if

then

. This proves that

converges to

Let

where

. Prove that

We want to prove given any

there exists

such that

Now, since

for all

we have that

Thus, if

then

. Let

be arbitrary. Let

be such that

. Then

. Therefore, if

then

and therefore

This proves that

Does the sequence

defined by

converge?

A useful tool for proving convergence is the following.

Let

be a sequence and let

. Let

be a sequence of positive numbers such that

. Suppose that there exists

such that

Then

Let

be arbitrary. Since

, there exists

such that

for all

. Let

. Then, if

then

and

. Thus, if

then

Suppose that

. Prove that

We first note that

where

and since

then

. Now, by Bernoulli's inequality (Example 1.2.5) it holds that

for all

and therefore

Now since

then it follows by Theorem 3.1.9 that

Consider the sequence

. Prove that

We have that

Using the definition of the limit of a sequence, one can show that

and therefore

Notice that in the definition of the limit of a sequence, we wrote there exists a number

. Could there be more than one number

satisfying the definition of convergence of a sequence? Before we go any further, we prove that if a sequence converges then it has a unique limit.

A convergent sequence can have at most one limit.

Suppose that

and that

. Let

be arbitrary. Then there exists

such that

for all

and there exists

such that

for all

. Let

. Then for

it holds that

and also

and therefore

Hence,

for all

, and therefore by Theoreom 2.2.7 we conclude that

, that is,

The ultimate long-time behavior of a sequence will not change if we discard a finite number of terms of the sequence. To be precise, suppose that

is a sequence and let

, that is,

is the sequence obtained from

by discarding the first

terms of

. In this case, we will call

the -tail of

. The next theorem states, not surprisingly, that the convergence properties of

and

are the same.

Let

be a sequence and let

be the sequence obtained from

by discarding the first

terms of

, in other words,

. Then

converges to

if and only if

converges to

Exercises

Write the first three terms of the recursively defined sequence

for

Use the definition of the limit of a sequence to establish the following limits:

\mbox{}

Prove that if and only if .
Combining the previous result and Example 3.1.10, prove that if then .
Conclude that for any real number , if then .

Let

and assume that

Prove that for all .
Use Theorem 3.1.9 to show that .

Note: Do not use Example 3.1.10 to show that

Suppose that

is non-empty and bounded above and let

. Show that there exists a sequence

such that

for all

and

. Hint: If

then clearly

. Since

there exists

such that

. Example 3.1.6 is similar.

Let

be the sequence defined as

Using the definition of the limit of a sequence, find

Limit Theorems

Proving that a particular number

is the limit of a given sequence is usually not easy because there is no systematic way to determine a candidate limit

for a given arbitrary sequence. Instead, we are frequently interested in just knowing if a given sequence converges or not, and not so much on finding the actual limit. The theorems in this section help us do just that. We begin with a definition.

A sequence

is said to be bounded if there exists

such that

for all

Prove that

is bounded if and only if there exists numbers

and

such that

for all

A convergent sequence is bounded.

Suppose that

converges to

. Then there exists

such that

for all

. Let

and we note that

. Then for all

it holds that

. Indeed, if

then

and if

then

Thus, for all

it holds that

and this proves that

is bounded.

then

Follows by the inequality

(see Corollary 2.3.6). Indeed, for any given

there exists

such that

for all

and therefore

for all

The following theorem describes how the basic operations of arithmetic preserve convergence.

Suppose that

and

Then and .
Then .
If and then .

(i) By the triangle inequality

Let

. There exists

such that

for

and there exists

such that

for

. Let

. Then for

The proof for

is similar.
(ii) We have that

Now,

is bounded because it is convergent, and therefore

for all

for some

. By convergence of

and

, there exists

such that

and

for all

. Therefore, if

then

(iii) It is enough to prove that

and then use (ii). Now, since

and

then

is bounded below by some positive number, say

. Indeed,

and

. Thus,

for all

. Now,

For

, there exists

such that

for all

. Therefore, for

we have that

Suppose that

. Then

for any

The next theorem states that the limit of a convergent sequence of non-negative terms is non-negative.

Suppose that

. If

for all

then

We prove the contrapositive, that is, we prove that if

then there exists

such that

. Suppose then that

. Let

be such that

. Since

, there exists

such that

, and thus by transitivity we have

Suppose that

and

are convergent and suppose that there exists

such that

for all

. Then

Suppose for now that

, that is,

for all

. Consider the sequence

. Then

for all

and

is convergent since it is the difference of convergent sequences. By Theorem 3.2.7, we conclude that

. But

and therefore

, which is the same as

. If

, then we can apply the theorem to the

-tail of the sequences of

and

and the result follows.

Suppose that

and

. Then

We have that

. The sequence

is constant and converges to

. The sequence

converges to

. Therefore, by the previous theorem,

, or

Suppose that

for all

. Assume that

and also

. Then

is convergent and

Let

be arbitrary. There exists

such that

for all

and there exists

such that

for all

. Let

. Then if

then

Therefore, for

we have that

and thus

Some people call the Squeeze Theorem the Sandwich Theorem; we are not those people.

Let

and let

. Prove that

Let

be a sequence such that

for all

and suppose that

exists. If

then

converges and

Let

be such that

and set

. There exists

such that

for all

. Therefore, for all

we have that

Thus,

, and therefore

, and inductively for

it holds that

Hence, the tail of the sequence

given by

satisfies

Since

it follows that

and therefore

by the Squeeze theorem. This implies that

converges to

also.

Exercises

Use the Limit Theorems to prove that if

converges and

converges then

converges. Give an example of two sequences

and

such that both

and

diverge but

converges.

Is the sequence

convergent? Explain.

Let

and

be sequences in

. Suppose that

and that

is bounded. Prove that

Show that if

and

are sequences such that

and

are convergent, then

is convergent.

Give examples of the following:

Divergent sequences and such that converges.
Divergent sequences and such that diverges.
A divergent sequence and a convergent sequence such that converges.
A divergent sequence and a convergent sequence such that diverges.

Let

and

be sequences and suppose that

converges to

. Assume that for every

there exists

such that

for all

. Prove that

also converges to

Let

be a sequence and define a sequence

for

. Show that if

then

Let

be a convergent sequence with limit

. Let

be a polynomial. Use the Limit Theorems to prove that the sequence

defined by

is convergent and find the limit of

Apply the Limit Theorems to find the limits of the following sequences:

Let

be a sequence such that

for all

. Suppose that

and

. Let

Prove that

Let

be a sequence of positive numbers such that

. Show that

is not bounded and hence is not convergent.

Monotone Sequences

As we have seen, a convergent sequence is necessarily bounded, and it is straightforward to construct examples of sequences that are bounded but not convergent, for example,

. In this section, we prove the Monotone Convergence Theorem which says that a bounded sequence whose terms increase (or decrease) must necessarily converge.

Let

be a sequence.

We say that is increasing if for all .
We say that is decreasing if for all .
We say that is monotone if is either increasing or decreasing.

Prove that if

is increasing then

is bounded below. Similarly, prove that if

is decreasing then

is bounded above.

is bounded and monotone then

is convergent. In particular:

if is bounded above and increasing then
if is bounded below and decreasing then

Suppose that

is bounded above and increasing. Let

and let

be arbitrary. Then by the properties of the supremum, there exists

such that

. Since

is increasing, and

is an upper bound for the range of the sequence, it follows that

for all

. Therefore,

for all

. Clearly, this implies that

for all

. Since

was arbitrary, this proves that

converges to

.
Suppose now that

is bounded below and decreasing. Let

and let

be arbitrary. Then by the properties of the infimum, there exists

such that

. Since

is decreasing, and

is a lower bound for the range of the sequence, it follows that

for all

. Therefore,

for all

. Hence,

for all

. Since

was arbitrary, this proves that

converges to

The Monotone Convergence Theorem (MCT) is an important tool in real analysis and we will use it frequently; notice that it is more-or-less a direct consequence of the Completeness Axiom. In fact, we could have taken as our starting axiom the MCT and then proved the Completeness property of

By the MCT, a bounded sequence that is also monotone is convergent. However, it is easy to construct a convergent sequence that is not monotone. Provide such an example.

The MCT can be used to show convergence of recursively defined sequences. To see how, suppose that

is defined recursively as

and

where

is some given function. For example, say

and

Hence, in this case

. If

is bounded and increasing then by the MCT

converges, but we do not know what the limit is. However, for example, if

is a polynomial/rational function of

then we can conclude that

must satisfy the equation

Indeed, if

is a polynomial/rational function then by the Limit Laws we have

But

and therefore

, which is equivalent to

since

is just the

-tail of the sequence

. Therefore,

as claimed. From the equation

we can solve for

if possible.

Consider the sequence

defined recursively as

and

Prove that

converges and find the limit.

We prove by induction that

for all

, that is,

is bounded below by

. First of all, it is clear that

. Now assume that

for some

. Then

Hence,

is bounded below by

. We now prove that

is decreasing. We compute that

, and thus

. Assume now that

for some

. Then

Hence, by induction we have shown that

is decreasing. By the MCT,

is convergent. Suppose that

. Then also

and the sequence

converges to

. Therefore,

and thus

Before we embark on the next example, we recall that

and if

then

and therefore

Consider the sequence

Note that this can be defined recursively as

and

. Prove that

converges.

We will prove by the MCT that

converges. By induction, one can show that

for all

. Therefore,

Hence,

and therefore

is bounded. Now, since

then clearly

. Thus

is increasing. By the MCT,

converges. You might recognize that

Let

and let

for

. Prove that

converges and find its limit.

Clearly,

. Assume by induction that

for some

. Then

Hence,

is an increasing sequence. We now prove that

is bounded above. Clearly,

. Assume that

for some

. Then

. This proves that

is bounded above. By the MCT,

converges, say to

. Moreover, since

(as can be proved by induction), then

. Therefore,

and then

. Hence,

. Since

then

Consider the sequence

. We will show that

is bounded and increasing, and therefore by the MCT

convergent. The limit of this sequence is the number

. From the binomial theorem

where we used that

for all

. This shows that

is bounded. Now, for each

, we have that

And similarly,

It is clear that

for all

. Hence,

. Therefore,

, that is,

is increasing. By the MCT,

converges to

Exercises

Let

be an increasing sequence, let

be a decreasing sequence, and assume that

for all

. Prove that

and

exist, and that

. {Note: Recall that a sequence

is bounded if there exist constants

(independent of ) such that

for all

Let

and let

for

. Prove that

is bounded and monotone. Find the limit of

Let

and let

for

. Prove that

is bounded and monotone. Find the limit of

. {Hint: Using induction to prove that

is monotone will not work with this sequence. Instead, work with

directly to prove that

is monotone. }

True or false, a convergent sequence is necessarily monotone? If it is true, prove it. If it is false, give an example.

Bolzano-Weierstrass Theorem

We can gather information about a sequence by studying its subsequences. Loosely speaking, a subsequence of

is a new sequence

such that each term

is from the original sequence

and the term

appears to the right of the term

in the original sequence

. Let us be precise about what we mean to the right.

Let

be a sequence. A subsequence of

is a sequence of the form

where

is a sequence of strictly increasing natural numbers. A subsequence of

will be denoted by

The notation

of a subsequence indicates that the indexing variable is

. The selection of the elements of

to form a subsequence

does not need to follow any particular well-defined pattern but only that

. Notice that for any increasing sequence

of natural numbers, we have

for all

An example of a subsequence of

is the sequence

. Here we have chosen the odd terms of the sequence

to create

. In other words, if we write that

then

. Another example of a subsequence of

is obtained by taking the even terms to get the subsequence

, so that here

. In general, we can take any increasing selection

, such as

to form a subsequence of

Two subsequences of

and

. Both of these subsequences converge to distinct limits.

We proved that

is countable and hence there is a bijection

. The bijection

defines a sequence

where

. Let

be arbitrary. By the density of

, there exists

such that

. Now consider the interval

. It has infinitely many distinct rational numbers (by the Density Theorem). Therefore, there exists

such that

. Consider now the interval

. It has infinitely many rational numbers, and therefore there exists

such that

. By induction, there exists a subsequence

such that

for all

. Therefore,

. We proved the following: For any real number

there exists a sequence of rational numbers that converges to

The following theorem is a necessary condition for convergence.

then every subsequence of

converges to

Let

. Then there exists

such that

for all

. Since

and

, then

for all

The contrapositive of the previous theorem is worth stating.

Let

be a sequence.

If has two subsequences converging to distinct limits then is divergent.
If has a subsequence that diverges then diverges.

The following is a very neat result that will supply us with a very short proof of the main result of this section, namely, the Bolzano-Weierstrass Theorem.

Every sequence has a monotone subsequence.

Let

be an arbitrary sequence. We will say that the term

is a peak if

for all

. In other words,

is a peak if it is an upper bound of all the terms coming after it. There are two possible cases for

, either it has an infinite number of peaks or it has a finite number of peaks. Suppose that it has an infinite number of peaks, say

, and we may assume that

. Then,

, and therefore

is a decreasing sequence. Now suppose that there are only a finite number of peaks and that

is the last peak. Then

is not a peak and therefore there exists

such that

. Similarly,

is not a peak and therefore there exists

such that

. Hence, by induction, there exists a subsequence

that is increasing.

Every bounded sequence contains a convergent subsequence.

Let

be an arbitrary bounded sequence. By Theorem 3.4.7,

has a monotone subsequence

. Since

is bounded then so is

. By the MCT applied to

we conclude that

is convergent.

We will give a second proof of the Bolzano-Weierstrass Theorem that is more hands-on.

is a bounded sequence, then there exists

such that

for all

. We will apply a recursive bisection algorithm to hunt down a converging subsequence of

. Let

be the mid-point of the interval

. Then at least one of the subsets

is infinite; if it is

then choose some

and let

and

; otherwise choose some

and let

. In any case, it is clear that

, that

and that

. Now let

be the mid-point of the interval

and let

. If

is infinite then choose some

and let

; otherwise choose some

and let

and

. In any case, it is clear that

, that

and

. By induction, there exists sequences

, and

such that

and

is increasing and

is decreasing. It is clear that

and

for all

. Hence, by the MCT,

and

are convergent. Moreover, since

then

and consequently

. By the Squeeze theorem we conclude that

Notice that the proofs of the Bolzano-Weierstrass Theorem rely on the Monotone Convergence Theorem and the latter relies on the Completeness Axiom. We therefore have the following chain of implications:

It turns out that if we had taken as our starting axiom the Bolzano-Weierstrass theorem then we could prove the Completeness property and then of course the MCT. In other words, all three statements are equivalent:

Exercises

Prove that the following sequences are divergent.

(Hint: Theorem 3.4.6)

Suppose that

for all

and suppose that

exists. Prove that

and that also

. (Hint: Consider subsequences of

Let

be a sequence.

Suppose that is increasing. Prove that if has a subsequence that is bounded above then is also bounded above.
Suppose that is decreasing. Prove that if has a subsequence that is bounded below then is also bounded below.

True or false: If

is bounded and diverges then

has two subsequences that converge to distinct limits. Explain.

Give an example of a sequence

with the following property: For each number

there exists a subsequence

such that

. Hint: If you are spending a lot of time on this question then you have not been reading this textbook carefully.

Suppose that

and

satisfy

and define

for

. Prove that

and

are convergent and that

. (Hint: First show that

for any

Let

be a sequence and define the sequence

Prove that if

then

limsup and liminf

The behavior of a convergent sequence is easy to understand. Indeed, if

then eventually the terms of

will be arbitrarily close to

for

sufficiently large. What else is there to say? In this section, we focus on bounded sequences that do not necessarily converge. The idea is that we would like to develop a limit concept for these sequences, and in particular, a limiting upper bound. Let

be an arbitrary sequence and introduce the set

defined as the set of all the limits of convergent subsequences of

, that is,

We will call

the subsequences limit set of

is bounded and

is the subsequences limit set of

explain why

is non-empty.

Here are six examples of sequences and the corresponding subsequences limit set.







enumeration of

Limits of subsequences

Notice that in the cases where

is bounded, the set

is also bounded, which is as expected since if

then for any convergent subsequence

we necessarily have

and therefore

In general, we have seen that for a general set

and

are not necessarily in

. This, however, is not the case for the subsequences limit set.

Let

be a bounded sequence and let

be its subsequences limit set. Then

and

. In other words, there exists a subsequence

such that

and similarly there exists a subsequence

such that

Let

. If

then there exists

such that

. Since

, there exists a subsequence

that converges to

. Therefore, there exists

such that

for all

. Hence, for each

, the inequality

holds for infinitely many

. Consider

. Then there exists

such that

. Now take

. Since

holds for infinitely many

, there exists

such that

and

. By induction, for

, there exists

such that

and

. Hence, the subsequence

satisfies

and therefore

. Therefore,

By Lemma 3.5.3, if

is a bounded sequence then there exists a convergent subsequence of

whose limit is larger than any other limit of a convergent subsequence of

. This leads to the following definition.

Let

be a bounded sequence and let

be its subsequences limit set. We define the limit superior of

and the limit inferior of

By Lemma 3.5.3,

is simply the largest limit of all convergent subsequences of

while

is the smallest limit of all convergent subsequences of

. Notice that by definition it is clear that

. The next theorem gives an alternative characterization of

and

. The idea is that

is a sort of limiting supremum and

is a sort of limiting infimum of a bounded sequence

Let

be a bounded sequence and let

. The following are equivalent:

If then there are at most finitely many such that and infinitely many such that .
Let . Then .

(i)

(ii) Let

denote the subsequences limit set of

. By definition,

and by Lemma 3.5.3 we have that

. Hence, there exists a subsequence of

converging to

and thus

holds for infinitely many

. In particular

holds for infinitely many

. Suppose that

holds infinitely often. Now

for all

and some

. Since the inequality

holds infinitely often, there exists a sequence

such that

for all

. We can assume that

is convergent (because it is bounded and we can pass to a subsequence by the MCT) and thus

. Hence we have proved that the subsequence

converges to a number greater than

which contradicts the definition of

.
(ii)

(iii) Let

. Since

holds for finitely many

, there exists

such that

for all

. Hence,

is an upper bound of

and thus

. Since

is decreasing, we have that

for all

. Now,

holds infinitely often and thus

for all

. Hence,

for all

. This proves the claim.
(iii)

(i) Let

be a convergent subsequence. Since

, by definition of

, we have that

. Therefore,

. Hence,

is an upper bound of

. By definition of

, there exists

such that

. By induction, there exists a subsequence

such that

. Hence, by the Squeeze Theorem,

. Hence,

and thus

Let

denote the

th prime number, that is

, and so on. The numbers

and

are called twin primes if

. The Twin Prime Conjecture is that

In other words, the Twin Prime Conjecture is that there are infinitely many pairs of twin primes.

We end this section with the following interesting theorem that says that if the subsequences limit set of a bounded sequence

consists of a single number

then the sequence

also converges to

Let

be a bounded sequence and let

. If every convergent subsequence of

converges to

then

converges to

Suppose that

does not converge to

. Then, there exists

such that for every

there exists

such that

. Let

. Then there exists

such that

. Then there exists

such that

. By induction, there exists a subsequence

such that

for all

. Now

is bounded and therefore by Bolzano-Weierstrass has a convergent subsequence, say

, which is also a subsequence of

. By assumption,

converges to

, which contradicts that

for all

Another way to say Theorem 3.5.7 is that if

is bounded and

then

. The converse, by the way, has already been proved: if

then every subsequence of

converges to

and therefore

Exercises

Determine the

and

for each case:

Let

and

be bounded sequences. Let

be the sequence

. Show that

In other words, prove that

Let

and

be bounded sequences. Show that if

for all

then

Cauchy Sequences

Up until now, the Monotone Convergence theorem is our main tool for determining that a sequence converges without actually knowing what the the limit is. It is a general sufficient condition for convergence. In this section, we prove another groundbreaking general sufficient condition for convergence known as the Cauchy criterion. Roughly speaking, the idea is that if the terms of a sequence

become closer and closer to one another as

then the sequence ought to converge. A sequence whose terms become closer and closer to one another is called a Cauchy sequence.

A sequence

is said to be a Cauchy sequence if for every

there exists a natural number

such that if

then

In other words,

is a Cauchy sequence if the difference

is arbitrarily small provided that both

and

are sufficiently large.

Prove directly using the definition of a Cauchy sequence that if

and

are Cauchy sequences then the sequence

is a Cauchy sequence.

Not surprisingly, a convergent sequence is indeed a Cauchy sequence.

is convergent then it is a Cauchy sequence.

Suppose that

. Let

and let

be sufficiently large so that

for all

. If

then by the triangle inequality,

This proves that

is Cauchy.

A Cauchy sequence is bounded.

is a Cauchy sequence then

is bounded.

The proof is similar to the proof that a convergent sequence is bounded.

The sequence

is convergent if and only if

is a Cauchy sequence.

In Lemma 3.6.3 we already showed that if

converges then it is a Cauchy sequence. To prove the converse, suppose that

is a Cauchy sequence. By Lemma 3.6.4,

is bounded. Therefore, by the Bolzano-Weierstrass theorem there is a subsequence

that converges, say it converges to

. We will prove that

also converges to

. Let

be arbitrary. Since

is Cauchy there exists

such that if

then

. On the other hand, since

there exists

such that

. Therefore, if

then

This proves that

converges to

Let

and suppose that

for all

. Using the fact that

for all

prove that if

then

. Deduce that

is a Cauchy sequence.

When the MCT is not applicable, the Cauchy criterion is another possible tool to show convergence of a sequence.

Consider the sequence

defined by

, and

for

. One can show that

is not monotone and therefore the MCT is not applicable.

Prove that for all .
Prove that for all .
Prove that if then Hint: Use part (b) and the Triangle inequality.
Deduce that is a Cauchy sequence and thus convergent.
Show by induction that and deduce that .

Notice that the main result used in the Cauchy Criterion is the Bolzano--Weierstrass (B--W) theorem. We therefore have the following chain of implications:

A close inspection of the Cauchy Criterion reveals that it is really a statement about the real numbers not having any gaps or holes. In fact, the same can be said about the MCT and the Bolzano-Weierstrass theorem. Regarding the Cauchy Criterion, if

is a Cauchy sequence then the terms of

are clustering around a number and that number must be in

has no holes. It is natural to ask then if we could have used the Cauchy Criterion as our starting axiom (instead of the Completeness Axiom) and then prove the Completeness property, and then the MCT and the Bolzano-Weierstrass theorem. Unfortunately, the Cauchy Criterion is not enough and we also need to take as an axiom the Archimedean Property.

Suppose that every Cauchy sequence in

converges to a number in

and the Archimedean Property holds in

. Then

satisfies the Completeness property, that is, every non-empty bounded above subset of

has a least upper bound in

Let

be a non-empty set that is bounded above. If

is an upper bound of

and

then

is the least upper bound of

and since

there is nothing to prove. Suppose then that no upper bound of

is an element of

. Let

be arbitrary and let

be an upper bound of

. Then

, and we set

. Consider the mid-point

of the interval

. If

is an upper bound of

then set

and set

, otherwise set

and

. In any case, we have

, and

. Now consider the mid-point

of the interval

. If

is an upper bound of

then set

and

, otherwise set

and

. In any case, we have

, and

. By induction, there exists a sequence

such that

is not an upper bound of

and a sequence

such that

is an upper bound of

, and

. By Exercise 3.6.6, and using the fact that

(this is where the Archimedean property is needed), it follows that

and

are Cauchy sequences and therefore by assumption both

and

are convergent. Since

it follows that

. We claim that

is the least upper bound of

. First of all, for fixed

we have that

for all

and therefore

, that is,

is an upper bound of

. Since

is not an upper bound of

, there exists

such that

and therefore by the Squeeze theorem we have

. Given an arbitrary

then there exists

such that

and thus

is not an upper bound of

. This proves that

is the least upper bound of

Exercises

Show that if

is a Cauchy sequence then

is bounded. (Note: Do not use the fact that a Cauchy sequence converges but show directly that if

is Cauchy then

is bounded.)

Show that if

converges then it is a Cauchy sequence.

Show by definition that

is a Cauchy sequence.

Show by definition that

is a Cauchy sequence.

Suppose that

and

are sequences such that

for all

. Show that if the sequence

is convergent then so is the sequence

Suppose that

. Show that if the sequence

satisfies

for all

then

is a Cauchy sequence and therefore convergent. Hint: If

then

Also, if

then

Infinite Series

Informally speaking, a series is an infinite sum:

Using summation notation:

The series

can be thought of as the sum of the sequence

. It is of course not possible to actually sum an infinite number of terms and so we need a precise way to talk about what it means for a series to have a finite value. Take for instance

so that the sequence being summed is

. Let's compute the first 10 terms of the sequence of partial sums

defined as follows:

With the help of a computer, one can compute

It seems as though the sequence

is converging to

, that is,

. It is then reasonable to say that the infinite series sums or converges to

We now introduce some definitions to formalize our example.

Let

be a sequence. The infinite series generated by

is the sequence

defined by

or recursively,

The sequence

is also called the sequence of partials sums generated by

. The

th term of the sequence of partial sums

can instead be written using summation notation:

Let

be the sequence

. The first few terms of the sequence of partials

In both examples above, we make the following important observation: if

is a sequence of non-negative terms then the sequence of partials sums

is increasing. Indeed, if

then

and thus

Find the sequence of partial sums generated by

We compute:

Hence,

The limit of the sequence

, if it exists, makes precise what it means for an infinite series

to converge.

Let

be a sequence and let

be the sequence of partial sums generated by

. If

exists and equals

then we say that the series generated by

converges to and we write that

The notation

is therefore a compact way of writing

and the question of whether the series

converges is really about whether the limit

exists. Often we will write a series such as

simply as

when either the initial value of

is understood or is unimportant. Sometimes, the initial

value may be

, or some other

The geometric series is perhaps the most important series we will encounter. Let

where

is a constant. The generated series is

and is called the geometric series. The

th term of the sequence of partial sums is

then

does not exist (why>, so suppose that

. Using the fact that

we can write

Now if

then

, while if

then

does not exist. Therefore, if

then

Therefore,

In summary: The series

is called the geometric series and converges if and only if

and in this case converges to

Consider the series

. The series can be written as

and thus it is a geometric series with

. Since

, the series converges and it converges to

Use the geometric series to show that

We can write

Consider

. Using partial fraction decomposition

Therefore, the

th term of the sequence of partial sums

This is a telescoping sum because all terms in the middle cancel and only the first and last remain. For example:

By induction one can show that

and therefore

. Therefore, the given series converges and it converges to

Consider the series

. We are going to analyze a subsequence

of the sequence of partial sums

. We will show that

is unbounded and thus

is unbounded and therefore divergent. Consequently, the series

is divergent. Consider

Now consider

Lastly consider,

In general, one an show by induction that for

we have

Therefore, the subsequence

is unbounded and thus the series

is divergent.

We now present some basic theorems on the convergence of series; most of them are a direct consequence of results from limit theorems for sequences. The first theorem we present can be used to show that a series diverges.

converges then

. Equivalently, if

then

diverges.

By definition, if

converges then the sequence of partial sums

is convergent. Suppose then that

. Recall that

has the recursive definition

. The sequences

and

both converge to

and thus

The series

is divergent because

The Series Divergence Test can only be used to show that a series diverges. For example, consider the harmonic series

. Clearly

. However, we already know that

is a divergent series. Hence, in general, the condition

is not sufficient to establish convergence of the series

A certain series

has sequence of partial sums

whose general

th term is

What is ?
Does the series converge? If yes, what does it converge to? Explain.
Does the sequence converge? If yes, what does it converge to? Explain.

The next theorem is just an application of the Cauchy criterion for convergence of sequences to the sequence of partial sums

The series

converges if and only if for every

there exists

such that if

then

The following theorem is very useful and is a direct application of the Monotone convergence theorem.

Suppose that

is a sequence of non-negative terms, that is,

for all

. Then

converges if and only if

is bounded. In this case,

Clearly, if

exists then

is bounded. Now suppose that

is bounded. Since

for all

then

and thus

shows that

is an increasing sequence. By the Monotone convergence theorem,

converges and

Consider the series

. Since

, to prove that the series converges it is enough to show that the sequence of partial sums

is bounded. We will consider a subsequence of

, namely,

where

for

. We have

and

By induction, one can show that

and therefore using the geometric series with

we have

This shows that the subsequence

is bounded. In general, the existence of a bounded subsequence does not imply that the original sequence is bounded but if the original sequence is increasing then it does. In this case,

is indeed increasing, and thus since

is a bounded subsequence then

is also bounded. Therefore,

converges, that is,

is a convergent series.

Suppose that

and

are convergent series.

Then and are also convergent and
For any constant , is also convergent and .

Using the limit laws:

Hence,

is a constant then by the Limit Laws,

Therefore,

Once establishing the convergence/divergence of some sequences, we can use comparison tests to the determine convergence/divergence properties of new sequences.

Let

and

be non-negative sequences and suppose that

for all

If converges then converges.
If diverges then diverges.

Let

and

be the sequences of partial sums. Since

then

. To prove (a), if

converges then

is bounded. Thus,

is also bounded. Since

is increasing and bounded it is convergent by the MCT. To prove (b), if

diverges then

is necessarily unbounded and thus

is also unbounded and therefore

is divergent.

Let

be an integer and let

be a sequence of integers such that

. Use the comparison test to show that the series

converges and converges to a point in

Determine whether the given series converge.

: First compute . Therefore, by the divergence test, the series diverges.
where : We know that is convergent. Now, if then . Therefore by the Comparison test, is also convergent. This is called a -series and actually converges for any .
: We will use the comparison test. We have The series converges and thus the original series also converges.

Suppose that

for all

. If

converges prove that

also converges. Does the claim hold if we only assume that

Since

then

. Since

converges then by the Comparison test then

also converges. More generally, suppose that

and

converges. Then

converges to zero and thus there exists

such that

for all

. Then since the series

converges it follows that

converges and consequently

converges.

The following two tests can be used for series whose terms are not necessarily non-negative.

If the series

converges then

converges.

From

we obtain that

converges then so does

. By the Comparison test 3.7.18, the series

converges also. Then the following series is a difference of two converging series and therefore converges:

and the proof is complete.

In the case that

converges we say that

converges absolutely. We end the section with the Ratio test for series.

Consider the series

and let

. If

then the series

converges absolutely and if

then the series diverges. If

or the limit does not exist then the test is inconclusive.

Suppose that

. Let

be such that

. There exists

such that

for all

and thus

for all

. By induction, it follows that

for all

. Since

is a geometric series with

then, by the comparison test, the series

converges. Therefore, the series

converges and by the Absolute convergence criterion we conclude that

converges absolutely. If

then a similar argument shows that

diverges. The case

follows from the fact that some series converge and some diverge when

Exercises

Suppose that

is a convergent series. Is it true that if

is divergent then

is divergent? If it is true, prove it, otherwise give an example to show that it is not true.

Suppose that

for all

. Prove that if

converges then

also converges. Is the converse true? That is, if

converges then does it necessarily follow that

converges?

Using only the tests derived in this section, determine whether the given series converge or diverge:

Suppose that

and

are non-negative sequences. Prove that if

and

are convergent then

is convergent. (Hint: Recall that if

then

converges iff

is bounded, where

is the sequence of partial sums. Alternatively, use the identity

Suppose that

for all

and

converges. You will prove that

converges.

Let . Prove that converges.
Using the fact that , prove that if . Deduce that
Deduce that converges.

Any number of the form

can be written as

Using this fact, and a geometric series, prove that

Show that the following series converge and find their sum.

Using the fact that

for all

, prove that the series

converges.

Let

be a decreasing sequence of non-negative terms. Let

be the sequence of partial sums of the series

, let

for

, and consider the subsequence

Show by induction that for all .
Conclude that if converges then .

(Note: This is a generalization of Example 3.7.16.)