2 Some Basic Techniques

Triangle Inequality

When we consider the limit of a sequence on the basis of the rigorous definition, we often use the triangle inequality explained in this section.

The absolute value of a real number x, denoted by |x|, is defined by
|x| = \left\{ \begin{array}{ll} x & ( x \geq 0 ) \\[2ex] -x & ( x < 0 ) . \end{array}\right.
That is, the absolute value |x| is the non-negative value of x without regard to its sign. From this definition, it is easy to see that | x | \geq 0 and | x | = 0 if and only if x = 0. In addition, we can easily see that |x| \geq x and |x|^2 = x^2. Moreover, the following properties hold:
| x y | = |x| \, |y|, \ \ \ \ \ | x + y | \leq |x| + |y|
The above inequality is called the triangle inequality, which is often used in arguments concerning the limit of sequences and functions. We can verify this inequality as follows:
\begin{array}{l} ( |x| + |y| )^2 - | x + y |^2 \\[2ex] \ \ \ \ = |x|^2 + 2|x| \, |y| + |y|^2 - ( x+ y )^2 \\[2ex] \ \ \ \ = |x|^2 + 2|x| \, |y| + |y|^2 - x^2 - 2xy \, - y^2 \\[2ex] \ \ \ \ = 2( |x| \, |y| - xy ) = 2( |xy| - xy ) \geq 0. \end{array}
We can calculate concrete expressions including the symbol of absolute values by using the above properties. Furthermore, it should be noted that the distance between x and y on the real line is given by | \, x \ - \ y \, |, which is a key point for understanding the meaning of results obtained by calculations using the symbol of absolute values.

 

In order to understand how to use the triangle inequality, we now show
\lim_{n \to \infty} a_n = \alpha, \ \ \lim_{n \to \infty} b_n = \beta \ \ \ \Longrightarrow \ \ \ \lim_{n \to \infty} ( a_n + b_n ) = \alpha + \beta,
which is a basic property of the limits of sequences.

Let c_n = a_n + b_n. We will show \displaystyle\lim_{n \to \infty} c_n = \alpha + \beta; that is,

For any \varepsilon > 0, there exists a natural number N(\varepsilon) such that for all n \geq N(\varepsilon), we have | \ c_n \ - \ (\alpha + \beta) \ | < \varepsilon.

This is an assertion to be proved. At present, the existence of N(\varepsilon), a function of \varepsilon, has not been guaranteed yet. On the other hand, \displaystyle\lim_{n \to \infty} a_n = \alpha and \displaystyle\lim_{n \to \infty} b_n = \beta are given conditions. Therefore, the following statements hold:

For any \varepsilon_1 > 0, there exists a natural number N_1(\varepsilon_1) such that for all n \geq N_1(\varepsilon_1), we have | \ a_n \ - \ \alpha \ | < \varepsilon_1,

and

For any \varepsilon_2 > 0, there exists a natural number N_2(\varepsilon_2) such that for all n \geq N_2(\varepsilon_2), we have | \ b_n \ - \ \beta \ | < \varepsilon_2.

Here, we use different notations to describe statements for each sequence, respectively, because \{ a_n \}, \{b_n \}, and \{ c_n \} are different sequences. The above statements for \{ a_n \} and \{b_n \} are given conditions, and the existence of N_1(\varepsilon_1), a function of \varepsilon_1, and that of N_2(\varepsilon_2), a function of \varepsilon_2, have already been guaranteed. Therefore, to prove the existence of N(\varepsilon), the only way to proceed is to define N(\varepsilon) by using the functions N_1(\varepsilon_1) and N_2(\varepsilon_2), whose existence has already been guaranteed.

To do so, we note the inequality | \ c_n \ - \ (\alpha + \beta) \ | < \varepsilon in the assertion to be proved. By using the triangle inequality, we have
\begin{array}{l} | \ c_n \ - \ (\alpha + \beta) \ | \\[2ex] \ \ \ \ = | \ ( a_n + b_n ) \ - \ ( \alpha + \beta ) \ | \\[2ex] \ \ \ \ = | \ ( a_n - \alpha ) \ + \ ( b_n - \beta ) \ | \\[2ex] \ \ \ \ \leq | \ a_n \ - \ \alpha \ | + | \ b_n \ - \ \beta \ |, \end{array}
and hence, for n \geq N_1(\varepsilon_1) and n \geq N_2(\varepsilon_2), we have
| \ c_n \ - \ ( \alpha + \beta ) \ | < \varepsilon_1 + \varepsilon_2.
Therefore, by setting \varepsilon_1 = \displaystyle\frac{\varepsilon}{2} and \varepsilon_2 = \displaystyle\frac{\varepsilon}{2}, when n is not less than both N_1(\displaystyle\frac{\varepsilon}{2}) and N_2(\displaystyle\frac{\varepsilon}{2}), we have
| \ c_n \ - \ (\alpha + \beta) \ | < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
Thus, for example, we define
N(\varepsilon) = \max{\large(} N_1(\displaystyle\frac{\varepsilon}{2}), N_2(\displaystyle\frac{\varepsilon}{2}) {\large)}, where \max( a, b) denotes the maximum of a and b. Then, noting that n satisfying n \geq N(\varepsilon) is not less than both N_1(\displaystyle\frac{\varepsilon}{2}) and N_2(\displaystyle\frac{\varepsilon}{2}), we see that for all n \geq N(\varepsilon),
| \ c_n \ - \ (\alpha + \beta) \ | < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
holds. That is, for all n \geq N(\varepsilon), we have | \ c_n \ - \ (\alpha + \beta) \ | < \varepsilon. It should be noted that the existence of the function N(\varepsilon) is really guaranteed because it is defined by the functions N_1(\varepsilon_1) and N_2(\varepsilon_2), whose existence has already been guaranteed.

Is there any other way to define N( \varepsilon)?

Yes, for example, we can define it as N(\varepsilon) = N_1(\displaystyle\frac{\varepsilon}{2}) + N_2(\displaystyle\frac{\varepsilon}{2}).

 

Exercise 2  Show that \displaystyle\lim_{n \to \infty} c a_n = c \alpha if \displaystyle\lim_{n \to \infty} a_n = \alpha, where c is a constant.

 

Logic Symbols

We often encounter the phrases “for any \cdots,” “for all \cdots,” and “there exists \cdots” in statements concerning the limits of sequences and functions. Therefore, noting the capital letter “A” of the words “any” and “all”, and reversing this letter vertically, we introduce the symbol “ \forall ” which means “any” and “all.” Similarly, noting the capital letter “E” of the word “exist,” and reversing this letter horizontally, we introduce the symbol “ \exists,” which means “exist.” These symbols enable us to simplify the description of statements concerning the limits of sequences and functions.

By using the symbols \forall and \exists, we can express the definition of the limit of a sequence “for any \varepsilon > 0 there exists a natural number N(\varepsilon) such that for all n \geq N(\varepsilon), we have | \ a_n \ - \ \alpha \ | < \varepsilon” as follows:

^{\forall}\varepsilon > 0, \ ^{\exists}N(\varepsilon) \in {\bf N}  s.t. ^{\forall}n \geq N(\varepsilon), \ | \ a_n \ - \ \alpha \ | < \varepsilon,

where {\bf N} denotes the set of natural numbers; N(\varepsilon) \in {\bf N} means that N(\varepsilon) is an element of {\bf N}; i.e., N(\varepsilon) is a natural number. Moreover, s.t. is the abbreviation for “such that.” Although this simplification saves time and effort in describing the definition of the limit of a sequence, we may find difficulty in using this simplification while we are unfamilar with rigorous arguments concerning the limits of sequences. Therefore, it is necessary for us to mentally read the above simplification as “for any \varepsilon > 0 there exists a natural number N(\varepsilon) such that for all n \geq N(\varepsilon), we have | \ a_n \ - \ \alpha \ | < \varepsilon.”

 

In order to get used to the above simplification using logic symbols such as \forall and \exists, we consider the following simple example.

Exmaple 1 : Show that \{ a_n \} is bounded if \{ a_n \} is convergent. Here, we say that \{ a_n \} is bounded if there exists a real number M > 0 such that for all n, we have | \, a_n \, | \leq M; i.e.,

^{\exists}M > 0 \ \ s.t. \ ^{\forall}n \in {\bf N},  \ | \, a_n \, | \leq M.

[Solution]  Since \{ a_n \} is convergent, for any \varepsilon > 0, there exists a natural number N(\varepsilon) such that for all n \geq N(\varepsilon), we have | \ a_n \ - \ \alpha \ | < \varepsilon; i.e.,

^{\forall}\varepsilon > 0, \ ^{\exists}N(\varepsilon) \in {\bf N}  s.t. ^{\forall}n \geq N(\varepsilon), \ | \ a_n \ - \ \alpha \ | < \varepsilon.

Since \varepsilon is arbitrary, for example, we can take \varepsilon = 1. Then, there exists a natural number N(1) such that for all n \geq N(1), we have | \ a_n \ - \ \alpha \ | < 1; i.e.,

^{\exists}N(1) \in {\bf N}  s.t. ^{\forall}n \geq N(1), \ | \ a_n \ - \ \alpha \ | < 1.

By using the triangle inequality, it follows from the last inequality in this statement that

| \ a_n \ | = | \ a_n \ - \ \alpha \ + \ \alpha \ | \leq | \ a_n \ - \ \alpha \ | + | \alpha | < 1 + | \alpha |,

which leads to

^{\exists}N(1) \in {\bf N}  s.t. ^{\forall}n \geq N(1), \ | \ a_n \ | < 1 + | \alpha |.

This means that there exists a natural number N(1) such that for all n \geq N(1), we have | \ a_n \ | < 1 + | \alpha |. Therefore, noting that the number N(1)-1 occurs just before N(1), we compare the absolute values of a_1, \ a_2, \ \cdots, \ a_{N(1)-1}, and 1 + | \alpha |. Let M be the maximum of | a_1 |, \ | a_2 |, \ \cdots, \ | a_{N(1) - 1} |, \ 1 + | \alpha |; i.e.,

M = \max\{ \ | a_1 |, \ | a_2 |, \ \cdots, \ | a_{N(1) - 1} |, \ 1 + | \alpha | \ \}.

Then, we see that for all n, | a_n | is not less than M; i.e.,

^{\exists}M > 0 \ \ s.t. \ ^{\forall}n \in {\bf N}, \ | \, a_n \, | \leq M.

Thus, \{ a_n \} is bounded. This completes the proof.

Is the key point of this solution to take \varepsilon = 1?

Yes. For a function N(\varepsilon) whose existence is guaranteed, by taking an existing number 1 and setting \varepsilon = 1, we can obtain a number N(1) whose existence is guaranteed. As a result, we can obtain a real number M whose existence is really guaranteed. Although we set \varepsilon = 1 in this solution, you can set another value, for example, \varepsilon = 100. It is crucial to take an existing number and substitute it into a function whose existence is guaranteed. Therefore, solutions where the value of \varepsilon is not specified ( \varepsilon is still arbitrary) would be quite nonsensical.

Exercise 3:  Show that the number of a_n‘s with a_n = 0 is finite if \displaystyle\lim_{n \to \infty} a_n = \alpha \neq 0.

 

In the solution of Example 1, we added sentences that explain the meaning of the simplified statements using logic symbols. However, in the solution of the next example, we do not add such sentences. Therefore, you should mentally add sentences to understand the meaning of the simplified statements that use logic symbols.

Example 2:  Show that \ \displaystyle\lim_{n \to \infty} a_n b_n = \alpha \beta  if \displaystyle\lim_{n \to \infty} a_n = \alpha and \displaystyle\lim_{n \to \infty} b_n = \beta .

[Solution] Let c_n = a_n b_n. Then, we will show that

^{\forall}\varepsilon > 0, \ ^{\exists}N(\varepsilon) \in {\bf N}  s.t. ^{\forall}n \geq N(\varepsilon), \ | \ c_n \ - \ \alpha \beta \ | < \varepsilon.

This is an assertion to be proved, and the existence of a function N(\varepsilon) has not been guaranteed yet. In contrast, we have

^{\forall}\varepsilon_1 > 0, \ ^{\exists}N_1(\varepsilon_1) \in {\bf N}  s.t. ^{\forall}n \geq N_1(\varepsilon_1), \ | \ a_n \ - \ \alpha \ | < \varepsilon_1

and

^{\forall}\varepsilon_2 > 0, \ ^{\exists}N_2(\varepsilon_2) \in {\bf N}  s.t. ^{\forall}n \geq N_2(\varepsilon_2), \ | \ b_n \ - \ \beta \ | < \varepsilon_2.

These are given conditions, and the existence of functions N_1(\varepsilon_1) and N_2(\varepsilon_2) has already been guaranteed. Therefore, our goal is to define N(\varepsilon) by using N_1(\varepsilon_1) and N_2(\varepsilon_2).

Noting the last inequality | \ c_n \ - \ \alpha \beta \ | < \varepsilon in the assertion to be proved, we estimate | \ c_n \ - \ \alpha \beta \ | as follows:
\begin{array}{l} \ | \ c_n \ - \ \alpha \beta \ | \ = \ | \ a_n b_n \ - \ \alpha\beta \ | \\[2ex] \ \ \ \ \ = \ | \ a_n b_n \ - \ a_n \beta \ + \ a_n \beta \ - \ \alpha \beta \ | \\[2ex] \ \ \ \ \ \leq \ | \ a_n ( b_n \ - \ \beta ) \ | \ + \ | \ ( a_n \ - \ \alpha ) \beta \ | \\[2ex] \ \ \ \ \ = | \ a_n \ | | \ b_n \ - \ \beta \ | \ + \ | \ a_n \ - \ \alpha \ | | \ \beta \ |. \end{array}
On the other hand, from Example 1, we see that \{ a_n \} is bounded; i.e.,

^{\exists}M > 0 \ \ s.t. \ ^{\forall}n \in {\bf N}, \ | \, a_n \, | \leq M.

Therefore, for n \geq N_1(\varepsilon_1) and n \geq N_2(\varepsilon_2), we have

| \ c_n \ - \ \alpha \beta \ | \ \leq \ | \ a_n \ | | \ b_n \ - \ \beta \ | \ + \ | \ a_n \ - \ \alpha \ | | \ \beta \ | \ \leq \ M \varepsilon_2 \ + \ | \ \beta \ | \varepsilon_1.

Hence, setting \varepsilon_1 = \displaystyle\frac{\varepsilon}{M + |\beta|} and \varepsilon_2 = \displaystyle\frac{\varepsilon}{M + |\beta|}, for n \geq N_1(\displaystyle\frac{\varepsilon}{M + |\beta|}) and n \geq N_2(\displaystyle\frac{\varepsilon}{M + |\beta|}), we have
| \ c_n \ - \ \alpha \beta \ | \leq M \cdot \frac{\varepsilon}{M + |\beta|} + |\beta| \cdot \frac{\varepsilon}{M + |\beta|} = \varepsilon.
Thus, we define
N(\varepsilon) = \max {\large(} N_1(\displaystyle\frac{\varepsilon}{M + |\beta|}), \ N_2(\displaystyle\frac{\varepsilon}{M + |\beta|}) {\large)}.
Then, noting that n \geq N(\varepsilon) implies n \geq N_1(\displaystyle\frac{\varepsilon}{M + |\beta|}) and n \geq N_2(\displaystyle\frac{\varepsilon}{M + |\beta|}), we see that

^{\forall}\varepsilon > 0, \ ^{\exists}N(\varepsilon) \in {\bf N}  s.t. ^{\forall}n \geq N(\varepsilon), \ | \ c_n \ - \ \alpha \beta \ | < \varepsilon.

This completes the proof.

As far as I recall from the examples discussed up to now, it is a key point to first note the last inequality in an assertion to be proved.

Indeed. A rigorous argument for the limits of sequences results in the proof of inequalities. When we try to prove inequalities, we note the conclusion to be proved at first, and drive our consideration from the conclusion toward the given assumptions. This is the usual way to prove inequalities.

Exercise 4 Suppose \displaystyle\lim_{n \to \infty} a_n = \alpha \neq 0 and \displaystyle\lim_{n \to \infty} b_n = \beta.

(1)  Show that ^{\exists}N_1 \in {\bf N}  s.t. ^{\forall}n \geq N_1, \ \displaystyle\frac{1}{| a_n |} < \displaystyle\frac{2}{|\alpha|}  ( Hint: Consider a ball (interval) with radius \displaystyle\frac{|\alpha|}{2} and center \alpha ).

(2)  Show \displaystyle\lim_{n \to \infty} \displaystyle\frac{1}{a_n} = \displaystyle\frac{1}{\alpha}.

(3)  Show \displaystyle\lim_{n \to \infty} \displaystyle\frac{b_n}{a_n} = \displaystyle\frac{\beta}{\alpha}.

 

Negation of Statement

In this section, we consider the negation of a statement including phrases such as “for any \cdots” and “there exists \cdots.”

For any language, it is usual to read a sentence from beginning to end in order. Moreover, the meaning of a word or a phrase in a sentence depends on the words and phrases appearing before that word or phrase. Therefore, in principle, we cannot change the order of the words and phrases in a sentence. For example, consider the following two statements:

(I) For any n \in {\bf N}, there exists M > 0 such that we have | \, a_n \, | \leq M.

(II) There exists M > 0 such that for any n \in {\bf N}, we have | \, a_n \, | \leq M.

Both these statements mean that | \, a_n \, | \leq M holds. However, the meaning of (I) is completely different from that of (II) because the order of the phrases “for any n \in N” and “there exists M > 0” is different in (I) and (II). In the case of (I), a positive real number M appears after a natural number n. Therefore, noting the basic principle that the meaning of a word depends on the words appearing before the word, a positive real number M exists depending on a natural number n. That is, M is a function of n, and hence M should be denoted by M(n). On the other hand, in the case of (II), there is nothing before a positive real number M. Therefore, a positive real number M exists independently. That is, M is not a function of n.

For example, consider a sequence \{ a_n \} given by a_n = n. This sequence satisfies (I). In fact, for any n \in N, we take M defined by M(n) = n + 1. Then, we have | a_n | = n < n + 1 = M. However, \{ a_n \} does not satisfy (II).

On the other hand, consider a sequence \{ a_n \} given by a_n = \displaystyle\frac{1}{n}. This sequence satisfies (II). In fact, we take M = 1. Then, for all n \in N, we have | a_n | = \displaystyle\frac{1}{n} \leq 1 = M. It should be noted that M = 1 exists independently; i.e., M = 1 is not a function of n.

I think that any sequence \{ a_n \} satisfies (I) because we can take M defined by M(n) = | a_n | + 1. On the other hand, recalling Example 1, (II) is the exact definition of a bounded sequence \{ a_n \}.

You are completely right.

In my textbook, the definition of the limit of a sequence is given by “for any \varepsilon > 0, there exists a natural number N such that for all n \geq N, we have | \ a_n \ - \ \alpha \ | \ < \ \varepsilon.” I have just understood a reason why the number N is not represented as N(\varepsilon) in this definition.

Indeed. It is tacitly understood that N is a function of \varepsilon. However, I think that N(\varepsilon) is a clear expression that prevents confusion and misunderstanding.

 

As explained above, we have confirmed that in principle, we cannot change the order of the words and phrases in a statement. We now explain how to negate a statement including the phrases “for any \cdots” and “there exists \cdots.”

Consider the following two statements:

(A) There is a method such that a dead person can be revived.

(B) For any road, the destination is Rome.

I think that these sentences are not used in everyday language.

Certainly. Mathematical language and everyday language, like oil and water, do not mix well. In everyday language, statements A and B are expressed by “A method exists that can revive a dead person” and “Every road leads to Rome”, respectively.

These statements are false. In fact, revival of a dead person contradicts our common sense based on the sciences. Moreover, every road in Japan, which consists of some islands surrounded by the ocean, does not lead to Rome in Italy. Therefore, the negations of statements A and B are true. They are as follows:

(A’) For any method, a dead person cannot be revived.

(B’) There is a road such that it does not lead to Rome.

In everyday language, statements A’ and B’ are expressed by “No method exists that can revive a dead person” and “There is a road that does not lead to Rome”, respectively.

The above examples suggest that:

The negation of the statement “there exists \cdots such that --- ” is the statement “for any \cdots, not ---

and

The negation of the statement “for any \cdots, ---” is the statement “there exists \cdots such that not ---

It should be noted that when we negate a statement, “exist” and “any” change to “any” and “exist,” respectively; i.e., the logic symbols “ \exists” and “ \forall” change to “ \forall” and “ \exists,” respectively.

 

Hereafter, we consider the negation of a mathematical statement. First, we consider the negation of the following statements:

(I) For any n \in {\bf N}, there exists M > 0 such that we have | \, a_n \, | \leq M,

and

(II) There exists M > 0 such that for any n \in {\bf N}, we have | \, a_n \, | \leq M,

which have been explained in the beginning of this section. Noting that the negation of | \, a_n \, | \leq M is given by | \, a_n \, | > M, we see that the negations of (I) and (II) are given by

(I’) There exists a natural number n \in {\bf N} such that for all M > 0, we have | \, a_n \, | > M,

and

(II’) For any M > 0, there exists a natural number n \in {\bf N} such that we have | \, a_n \, | > M,

respectively. Here, M is a function of n in statement I, but there is no relationship between M and n in statement I’. Similarly, there is no relationship between M and n in statement II, but n is a function of M in statement II’. We should take special care not to overlook the fact that such relationships appear or disappear when we negate a statement.

Exercise 5:  Is there a sequence satisfying (I’) ? How about (II’) ?

 

We now consider the negation of the definition of the limit of a sequence: “For any \varepsilon > 0, there exists a natural number N(\varepsilon) such that for all n \geq N(\varepsilon), we have | \ a_n \ - \ \alpha \ | < \varepsilon.” The statement of this definition consists of two blocks as follows:

(P)  For any \varepsilon > 0, there exists a natural number N(\varepsilon) such that the following condition (A) holds.

(C)  (A) : For all n \geq N(\varepsilon), we have | \ a_n \ - \ \alpha \ | < \varepsilon.

Here, (P) and (C) indicate “parent” and “child,” respectively. First, we consider the negation of (P). It is given by:

(P’)  There exists a real number \varepsilon > 0 such that for any N, the following condition (A) does not hold.

It should be noted that the relationship between N and \varepsilon is lost in (P’). Next, we consider the negation of (C). It is given by:

(C’)   (A’) : There exists a natural number n \geq N such that we have | \ a_n \ - \ \alpha \ | \geq \varepsilon.

Third, changing “the following condition (A) does not hold” to “the following condition (A’) holds” in (P’), we have:

(P’)  There exists a real number \varepsilon > 0 such that for any N, the following condition (A’) holds.

(C’)  (A’) : There exists a natural number n(N) \geq N such that we have | \ a_{n(N)} \ - \ \alpha \ | \geq \varepsilon.

It should be noted that the relationship between n and N appears; i.e., n is a function of N. Finally, combining (P’) with (C’), we obtain the following:

There exists a real number \varepsilon > 0 such that for any N, there exists a natural number n(N) \geq N such that we have | \ a_{n(N)} \ - \ \alpha \ | \geq \varepsilon.

This is the negation of the definition of the limit of a sequence, which means that \alpha is not the limit value of \{ a_n \}. In order to understand the meaning of this statement, we examine the limit of the sequence
-1, \ 1, -1, \ 1, -1, \ \cdots,
which is given by a_n = (-1)^n.

First, we will show that \alpha = -1 is not the limit value of \{ a_n \}. Let \varepsilon = 1. For any N, we define a natural number n by n(N) = 2N. Then, we have n(N) = 2N \geq N, which means that n(N) is not less than N. Moreover, we have
| \ a_{n} \ - \ \alpha \ | = | \ a_{2N} \ - \ (-1) \ | = | \ (-1)^{2N} \ + \ 1 \ | = 1 + 1 = 2 > 1,
which implies | \ a_{n(N)} \ - \ \alpha \ | > \varepsilon. Therefore, there exists a real positive number \varepsilon = 1 such that for any N, there exists a natural number n(N) \geq N such that we have | \ a_{n(N)} \ - \ \alpha \ | > \varepsilon. Hence, \alpha = -1 is not the limit value. Similarly, we can show that \alpha = 1 is not the limit value of \{ a_n \}.

Exercise 6   Show that \alpha is not the limit value of \{ a_n \} if | \alpha | > 1.

 

Finally, we will show that \alpha is not the limit value of \{ a_n \} if | \alpha | < 1. Let \varepsilon = \displaystyle\frac{1 \ - \ | \alpha |}{2} > 0. For any N, we define a natural number n by n(N) = N. Then, we have n(N) = N \geq N, which means that n(N) is not less than N. Moreover, we have
| \ a_n \ - \ \alpha \ | = | \ (-1)^n \ - \ \alpha \ | \geq 1 \ - \ |\alpha| > \frac{1 \ - \ |\alpha|}{2}.
Here, we have used the inequality | \, x \ - \ y \, | \geq |x| \ - \ |y|, which is derived from the triangle inequality. In fact, noting that
| x | = | \, x \ - \ y \ + \ y \, | \leq | \, x \ - \ y \, | + | y |,
we have | x | \leq | \, x \ -\ y \, | + | y |, which implies | x | \ - \ | y | \leq | \, x \ - \ y \, |. Therefore, we have
| \ a_{n(N)} \ - \ \alpha \ | > \varepsilon.
As a result, when | \alpha | < 1, there exists a positive real number \varepsilon = \displaystyle\frac{1 \ - \ | \alpha |}{2} such that for any N, there exists a natural number n(N) \geq N such that we have | \ a_{n(N)} \ - \ \alpha \ | > \varepsilon. Hence, \alpha is not the limit value of \{ a_n \} if | \alpha | < 1.

Thus, we see that no \alpha exists that is the limit value of \{ a_n \}. This implies that \{ a_n \} is not convergent.

As seen in the above example, in order to show that a sequence is not convergent, we have to show that no \alpha exists that is the limit value of the sequence. I think that this is rather tedious.

Indeed. “For any \varepsilon > 0, there exists a natural number N(\varepsilon) such that for all n \geq N(\varepsilon), we have | \ a_n \ - \ \alpha \ | < \varepsilon” is the definition of the limit value of \{ a_n \}, \alpha. It does not define that \{ a_n \} is convergent.

If so, how can we define a convergent sequence?

We add a phrase “There exists a real number \alpha” to “for any \varepsilon > 0, there exists a natural number N(\varepsilon) such that for all n \geq N(\varepsilon), we have | \ a_n \ - \ \alpha \ | < \varepsilon.”

Then, the definition of a convergent sequence is given by “There exists a real number \alpha such that for any \varepsilon > 0, there exists a natural number N(\varepsilon) such that for all n \geq N(\varepsilon), we have | \ a_n \ - \ \alpha \ | < \varepsilon.” This is lengthy and not easy to read. However, I understand why we must consider all cases of \alpha to show that a sequence is not convergent; “for any \alpha” is the negation of “there exists a real number \alpha.”

Many textbooks give the definition of the limit of a sequence by “for any \varepsilon > 0, there exists a natural number N(\varepsilon) such that for all n \geq N(\varepsilon), we have | \ a_n \ - \ \alpha \ | < \varepsilon.” This is not the definition of a convergent sequence, but the definition of the limit value, \alpha. As long as we use the lenghty definition of a convergent sequence, we can prove the convergence of a sequence if and only if we can predict the limit value of the sequence.

If so, how can we prove that a sequence is convergent if we cannot predict the limit value of the sequence? In addition, when a sequence is not convergent, do we have to examine all cases of \alpha in the same way as for the sequence (-1)^n?

Actually, there is a way of examining the convergence of a sequence if we cannot predict its limit value. The following condition enables us to prove the convergence of a sequence: “For any \varepsilon > 0, there exists a natural number N(\varepsilon) such that for all m, \ n \geq N(\varepsilon), we have | \ a_m \ - \ a_n \ | < \varepsilon,” which is called the Cauchy condition. We can apply this condition if we cannot predict the limit value, because it does not appear in the Cauchy condition. In fact, by using the Cauchy condition, we can easily find that (-1)^n is not convergent.

I see. However, how can we prove that a sequence satisfying the Cauchy condition is really convergent?

It is rather difficult, and we do not have enough time to explain the proof. You should consult advanced textbooks of calculus.

 

Exercise 7:  Let a_n be the remainder obtained by dividing n by 3. Show that \{ a_n \} is not convergent.

Good job! Let’s go to the next section.

 

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