Monotonic number sequences are. I

Purpose: To give the concept, definition of a sequence, finite, infinite, various ways of defining sequences, their differences, teach how to use them when solving examples.

Equipment: Tables.

Progress of the lesson

I. Organizational moment.

II. Frontal check homework:

1) student on the board task No. 2.636 (from part II of the “Collection of tasks for the written exam in grade 9)

2) student. Build a graph

3) frontally with the entire class No. 2.334 (a).

III. Explanation of new material.

A school lecture is a form of organizing the educational process that orients students when studying a particular topic to the main thing and involves a broad demonstration of the personal attitude of the teacher and students to the educational material. Because The lesson-lecture provides for a large-block presentation of the material by the teacher, then verbal communication between the teacher and students is the main thing in its technology. The teacher’s word has an emotional, aesthetic impact and creates a certain attitude towards the subject. With the help of a lecture, various types of student activities in the classroom are guided, and through knowledge, skills and abilities, cognition is formed as the basis of educational activity.

I. Write down two-digit numbers ending in 3 in ascending order.

13; 23; 33;………….93.

To everyone serial number From 1 to 9, match a specific two-digit number:

1->13; 2->23;………9->93.

A correspondence has been established between the set of the first nine natural numbers and the set of two-digit numbers ending in 3. This correspondence is a function.

The domain of definition is (1; 2; 3;……..9)

Many values ​​(13; 23; 33;…….93).

If the correspondence is denoted by f, then

This sequence can be specified using par.

(1;3) (2;5) (3;7) (4;9)

b) 1; 0; 1; 0; 1; 0

Table No. 1

A)

b)

II.

O.o.f. (1; 2; 3; 4;…..)

M.z.f. g(1) = ;

g(3) =; ...

g(60) =

A function defined on the set of natural numbers is called an infinite sequence.

c) 2; 4; 6; 8; 10;……..

1 -> 2; 2 -> 4; ……. n -> 2n

f(1); f(2); f(3)… …..f(n)

- members of the sequence.

Note: it is necessary to distinguish between the concept of a set and the concept of a sequence.

a) (10; 20; 30; 40)

The same set.

{40; 30; 20; 10}

b) however, sequences 10; 20; 30; 40

Various:

(1; 10) (2;20) (3;30) (4;40)

(1; 40) (2;30) (3;20) (4;10).

III. Consider the sequence:

1) 3; 5; 7; 9; 11;……. -> infinite, increasing

2) 10; 9; 8; 7; 6. -> final, decreasing.

A)

A sequence is called increasing if each member, starting from the second, is greater than its previous one.

b)

The definition of a decreasing sequence is given.

Increasing or decreasing sequences are called monotonic.

1; 0; 1; 0; 1; 0. - fluctuating;

5; 5; 5; 5; ….. - constant.

IV. Sequences can be depicted geometrically. Because sequence is a function whose domain of definition is the set N, then the graph, apparently, is the set of points of the plane (x; y).

Example: -3; -2; -1; 0; 1; 2; 3.

(1; -3) (2;-2) (3; -1) (4; 0) (5;1) (6;2) (7;3)

Let's plot this sequence

Figure 1.

Example: Prove that a sequence given in this form

99; 74; 49; 24; -1;……………

is decreasing.

V. Methods for specifying sequences.

Because A sequence is a function defined on the set N, then there are five ways to define sequences:

I. Tabular

II. Description method

III. Analytical

IV. Graphic

V. Recurrent

I. Tabular - very inconvenient. We draw up a table and use it to determine which member? what place does he take……..

II. Method of description.

Example: The sequence is such that each member is written using the number 4, and the number of digits is equal to the number of the sequence.

III. Analytical method(using formula).

A formula that expresses each member of a sequence in terms of its number n is called a formula for the n member of the sequence.

For example:

and students make up these sequences, and vice versa: choose a formula for the terms of the sequences:

a) 1; ; ;…………..
b) ...
V)
G)
e) 1;-2;3;-4;5;-6;………….

IV. Graphic method- also not very convenient, they usually don’t use it.

Weierstrass's theorem on the limit of a monotone sequence

Any monotonic bounded sequence (xn) has final limit, equal to the exact upper boundary, sup(xn) for a non-decreasing and exact lower bound, inf(xn) for a non-increasing sequence.
Any monotonic unbounded sequence has infinite limit, equal to plus infinity for a non-decreasing sequence and minus infinity for a non-increasing sequence.

Proof

1) non-decreasing limited sequence .

(1.1) .

Since the sequence is bounded, it has a tight upper bound
.
This means that:

• for all n,
  (1.2) ;
• 
  (1.3) .

.
Here we also used (1.3). Combining with (1.2), we find:
at .
Since then
,
or
at .
The first part of the theorem has been proven.

2) Let now the sequence be non-increasing bounded sequence:
(2.1) for all n.

Since the sequence is bounded, it has a tight lower bound
.
This means the following:

• for all n the following inequalities hold:
  (2.2) ;
• for anyone positive number, there is a number, depending on ε, for which
  (2.3) .

.
Here we also used (2.3). Taking into account (2.2), we find:
at .
Since then
,
or
at .
This means that the number is the limit of the sequence.
The second part of the theorem is proven.

Now let's consider unlimited sequences.
3) Let the sequence be unlimited non-decreasing sequence.

Since the sequence is non-decreasing, the following inequalities hold for all n:
(3.1) .

Since the sequence is non-decreasing and unbounded, it is unbounded with right side. Then for any number M there is a number, depending on M, for which
(3.2) .

Since the sequence is non-decreasing, then when we have:
.
Here we also used (3.2).

.
This means that the limit of the sequence is plus infinity:
.
The third part of the theorem is proven.

4) Finally, consider the case when unbounded non-increasing sequence.

Similar to the previous one, since the sequence is non-increasing, then
(4.1) for all n.

Since the sequence is non-increasing and unbounded, it is unbounded on the left side. Then for any number M there is a number, depending on M, for which
(4.2) .

Since the sequence is non-increasing, then when we have:
.

So, for any number M there is a natural number depending on M, so that for all numbers the following inequalities hold:
.
This means that the limit of the sequence is minus infinity:
.
The theorem has been proven.

Example of problem solution

Using Weierstrass's theorem, prove sequence convergence:
, , . . . , , . . .
Then find its limit.

Let's represent the sequence in the form of recurrent formulas:
,
.

Let us prove that the given sequence is bounded above by the value
(P1) .
We carry out the proof using the method mathematical induction.
.
Let . Then
.
Inequality (A1) is proven.

Let us prove that the sequence increases monotonically.
;
(P2) .
Since , then the denominator of the fraction and the first factor in the numerator are positive. Due to the limitation of the terms of the sequence by inequality (A1), the second factor is also positive. That's why
.
That is, the sequence is strictly increasing.

Since the sequence is increasing and bounded above, it is a bounded sequence. Therefore, according to Weierstrass's theorem, it has a limit.

Let's find this limit. Let's denote it by a:
.
Let's use the fact that
.
Let's apply this to (A2), using the arithmetic properties of limits of convergent sequences:
.
The condition is satisfied by the root.

If everyone natural number n is assigned to some real number x n , then they say that it is given number sequence

x 1 , x 2 , … x n , …

Number x 1 is called a member of the sequence with number 1 or first term of the sequence, number x 2 - member of the sequence with number 2 or the second member of the sequence, etc. The number x n is called member of the sequence with number n.

There are two ways to specify number sequences - with and with recurrent formula.

Sequence using formulas general member sequences– this is a sequence task

x 1 , x 2 , … x n , …

using a formula expressing the dependence of the term x n on its number n.

Example 1. Number sequence

1, 4, 9, … n 2 , …

given using the common term formula

x n = n 2 , n = 1, 2, 3, …

Specifying a sequence using a formula expressing a sequence member x n through the sequence members with preceding numbers is called specifying a sequence using recurrent formula.

x 1 , x 2 , … x n , …

called in increasing sequence, more previous member.

In other words, for everyone n

x n + 1 >x n

Example 3. Sequence of natural numbers

1, 2, 3, … n, …

is ascending sequence.

Definition 2. Number sequence

x 1 , x 2 , … x n , …

called descending sequence if each member of this sequence less previous member.

In other words, for everyone n= 1, 2, 3, … the inequality is satisfied

x n + 1 < x n

Example 4. Subsequence

given by the formula

is descending sequence.

Example 5. Number sequence

1, - 1, 1, - 1, …

given by the formula

x n = (- 1) n , n = 1, 2, 3, …

is not neither increasing nor decreasing sequence.

Definition 3. Increasing and decreasing number sequences are called monotonic sequences.

Bounded and Unbounded Sequences

Definition 4. Number sequence

x 1 , x 2 , … x n , …

called limited from above, if there is a number M such that each member of this sequence less numbers M.

In other words, for everyone n= 1, 2, 3, … the inequality is satisfied

Definition 5. Number sequence

x 1 , x 2 , … x n , …

called bounded below, if there is a number m such that each member of this sequence more numbers m.

In other words, for everyone n= 1, 2, 3, … the inequality is satisfied

Definition 6. Number sequence

x 1 , x 2 , … x n , …

is called limited if it limited both above and below.

In other words, there are numbers M and m such that for all n= 1, 2, 3, … the inequality is satisfied

m< x n < M

Definition 7. Number sequences, which are not limited, called unlimited sequences.

Example 6. Number sequence

1, 4, 9, … n 2 , …

given by the formula

x n = n 2 , n = 1, 2, 3, … ,

bounded below, for example, the number 0. However, this sequence unlimited from above.

Example 7. Subsequence

given by the formula

is limited sequence, because for everyone n= 1, 2, 3, … the inequality is satisfied

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The monotony of the sequence

Monotonic sequence- sequence satisfying one of the following conditions:

Among the monotonic sequences, the following stand out: strictly monotonous sequences satisfying one of the following conditions:

Sometimes a variant of terminology is used in which the term "increasing sequence" is considered as a synonym for the term "non-decreasing sequence", and the term "decreasing sequence" is considered as a synonym for the term "non-increasing sequence". In such a case, the increasing and decreasing sequences from the above definition are called “strictly increasing” and “strictly decreasing”, respectively.

Some generalizations

It may turn out that the above conditions are not met for all numbers, but only for numbers from a certain range

(here it is allowed to reverse the right border N+ to infinity). In this case the sequence is called monotonic on the interval I , and the range itself I called an interval of monotony sequences.

Examples

See also

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Definition 1. The sequence is called decreasing (non-increasing ), if for everyone
inequality holds
.

Definition 2. Consistency
called increasing (non-decreasing ), if for everyone
inequality holds
.

Definition 3. Decreasing, non-increasing, increasing and non-decreasing sequences are called monotonous sequences, decreasing and increasing sequences are also called strictly monotonous sequences.

Obviously, a non-decreasing sequence is bounded from below, and a non-increasing sequence is bounded from above. Therefore, any monotonic sequence is obviously limited on one side.

Example 1. Consistency
increases, does not decrease,
decreases
does not increase
– non-monotonic sequence.

For monotonic sequences, the following plays an important role:

Theorem 1. If a nondecreasing (nonincreasing) sequence is bounded above (below), then it converges.

Proof. Let the sequence
does not decrease and is bounded from above, i.e.
and many
limited from above. By Theorem 1 § 2 there is
. Let's prove that
.

Let's take
arbitrarily. Because A– exact upper bound, there is a number N such that
. Since the sequence is non-decreasing, then for all
we have, i.e.
, That's why
for everyone
, and this means that
.

For a nonincreasing sequence bounded below, the proof is similar to ( students can prove this statement at home on their own). The theorem has been proven.

Comment. Theorem 1 can be formulated differently.

Theorem 2. In order for a monotonic sequence to converge, it is necessary and sufficient that it be bounded.

Sufficiency is established in Theorem 1, necessity – in Theorem 2 of § 5.

The monotonicity condition is not necessary for the convergence of a sequence, since a convergent sequence is not necessarily monotonic. For example, the sequence
not monotonic, but converges to zero.

Consequence. If the sequence
increases (decreases) and is limited from above (from below), then
(
).

Indeed, by Theorem 1
(
).

Definition 4. If
at
, then the sequence is called contracting system of nested segments .

Theorem 3 (principle of nested segments). Every contracting system of nested segments has, and moreover, a unique point With, belonging to all segments of this system.

Proof. Let us prove that the point With exists. Because
, That
and therefore the sequence
does not decrease, but the sequence
does not increase. At the same time
And
limited because. Then, by Theorem 1, there exist
And
, but since
, That
=
. Found point With belongs to all segments of the system, since by the corollary of Theorem 1
,
, i.e.
for all values n.

Let us now show that the point With- the only one. Let's assume that there are two such points: With And d and let for certainty
. Then the segment
belongs to all segments
, i.e.
for everyone n, which is impossible, since
and, therefore, starting from a certain number,
. The theorem has been proven.

Note that the essential thing here is that closed intervals are considered, i.e. segments. If we consider a system of contracting intervals, then the principle is, generally speaking, incorrect. For example, intervals
, obviously contract to a point
, however point
does not belong to any interval of this system.

Let us now consider examples of convergent monotonic sequences.

1) Number e.

Let us now consider the sequence
. How is she behaving? Base

degrees
, That's why
? On the other side,
, A
, That's why
? Or is there no limit?

To answer these questions, consider the auxiliary sequence
. Let us prove that it decreases and is bounded below. At the same time, we will need

Lemma. If
, then for all natural values n we have

(Bernoulli's inequality).

Proof. Let's use the method of mathematical induction.

If
, That
, i.e. the inequality is true.

Let's assume that it is true for
and prove its validity for
+1.

Right
. Let's multiply this inequality by
:

Thus, . This means, according to the principle of mathematical induction, Bernoulli’s inequality is true for all natural values n. The lemma is proven.

Let us show that the sequence
decreases. We have

‌‌‌׀Bernoulli's inequality׀
, and this means that the sequence
decreases.

Boundedness from below follows from the inequality
‌‌‌׀Bernoulli's inequality׀
for all natural values n.

By Theorem 1 there is
, which is denoted by the letter e. That's why
.

Number e irrational and transcendental, e= 2.718281828… . It is, as is known, the base of natural logarithms.

Notes. 1) Bernoulli's inequality can be used to prove that
at
. Indeed, if
, That
. Then, according to Bernoulli’s inequality, with
. Hence, at
we have
, that is
at
.

2) In the example discussed above, the base of the degree tends to 1, and the exponent n- To , that is, there is uncertainty of the form . Uncertainty of this kind, as we have shown, is revealed by the remarkable limit
.

2)
(*)

Let us prove that this sequence converges. To do this, we show that it is bounded from below and does not increase. In this case, we use the inequality
for everyone
, which is a consequence of the inequality
.

We have
see inequality is higher
, i.e. the sequence is bounded below by the number
.

Next,
since

, i.e. the sequence does not increase.

By Theorem 1 there is
, which we denote X. Passing in equality (*) to the limit at
, we get

, i.e.
, where
(we take the plus sign, since all terms of the sequence are positive).

The sequence (*) is used in the calculation
approximately. For take any positive number. For example, let's find
. Let
. Then
,. Thus,
.

3)
.

We have
. Because
at
, there is a number N, such that for everyone
inequality holds
. So the sequence
, starting from some number N, decreases and is bounded below, since
for all values n. This means that by Theorem 1 there is
. Because
, we have
.

So,
.

4)
, right – n roots.

Using the method of mathematical induction we will show that
for all values n. We have
. Let
. Then, from here we obtain a statement based on the principle of mathematical induction. Using this fact, we find, i.e. subsequence
increases and is bounded from above. Therefore it exists because
.

Thus,
.