What does monotonic sequence mean? Weierstrass's theorem on the limit of a monotone sequence. Bounded and Unbounded Sequences

The elements of which do not decrease with increasing number, or, conversely, do not increase. Such sequences are often encountered in research and have a number of distinctive features and additional properties. A sequence of one number cannot be considered ascending or descending.

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  Let there be a set X (\displaystyle X), on which the order relation is introduced.

  Sequence of set elements X (\displaystyle X) called non-decreasing , if each element of this sequence is not greater than the next one.

  ( x n ) (\displaystyle \(x_(n)\))- non-decreasing ⇔ ∀ n ∈ N: x n ⩽ x n + 1 (\displaystyle \Leftrightarrow ~\forall n\in \mathbb (N) \colon x_(n)\leqslant x_(n+1))

  Subsequence ( x n ) (\displaystyle \(x_(n)\)) elements of the set X (\displaystyle X) called non-increasing , if each next element of this sequence does not exceed the previous one.

  ( x n ) (\displaystyle \(x_(n)\))- non-increasing ⇔ ∀ n ∈ N: x n ⩾ x n + 1 (\displaystyle \Leftrightarrow ~\forall n\in \mathbb (N) \colon x_(n)\geqslant x_(n+1))

  Subsequence ( x n ) (\displaystyle \(x_(n)\)) elements of the set X (\displaystyle X) called increasing , if each next element of this sequence is greater than the previous one.

  ( x n ) (\displaystyle \(x_(n)\))- increasing ⇔ ∀ n ∈ N: x n< x n + 1 {\displaystyle \Leftrightarrow ~\forall n\in \mathbb {N} \colon x_{n}

  Subsequence ( x n ) (\displaystyle \(x_(n)\)) elements of the set X (\displaystyle X) called decreasing , if each element of this sequence is greater than the next one.

  ( x n ) (\displaystyle \(x_(n)\))- decreasing ⇔ ∀ n ∈ N: x n > x n + 1 (\displaystyle \Leftrightarrow ~\forall n\in \mathbb (N) \colon x_(n)>x_(n+1))

  monotonous, if it is non-decreasing or non-increasing.

  The sequence is called strictly monotonous, if it is increasing or decreasing.

  Obviously, a strictly monotonic sequence is monotonic.

  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.

  Intervals of monotony

  It may turn out that the above conditions are not met for all numbers n ∈ N (\displaystyle n\in \mathbb (N) ), but only for numbers from a certain range

  I = ( n ∈ N ∣ N − ⩽ n< N + } {\displaystyle I=\{n\in \mathbb {N} \mid N_{-}\leqslant n

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

  Weierstrass's theorem on the limit of a monotone sequence

  Any monotone bounded sequence (xn) has a finite limit equal to the exact upper limit, sup(xn) for a non-decreasing and exact lower bound, inf(xn) for a non-increasing sequence.
  Any monotonic unbounded sequence has an 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 finite upper bound
  .
  It means that:
  

  • for all n,
    (1.2) ;
  • for any positive number, there is a number depending on ε, so that
    (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 finite lower bound
  .
  This means the following:
  

  • for all n the following inequalities hold:
    (2.2) ;
  • for any 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 consider unbounded 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 on the 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 equal to minus infinity:
  .
  The theorem is proven.

  Example of problem solution

  Using Weierstrass's theorem, prove the convergence of the sequence:
  , , . . . , , . . .
  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 each natural number n is associated with some real number x n, then we say that the 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 for the general term of a sequence– 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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  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 the 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 all
  , and this means that
  .

  For a non-increasing sequence bounded below, the proof is similar to ( students can prove this statement at home on their own). The theorem is 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 the
  , That
  and therefore the sequence
  does not decrease, but the sequence
  does not increase. Wherein
  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 all n, which is impossible, since
  and, therefore, starting from a certain number,
  . The theorem is 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 all
  , which is a consequence of the inequality
  .

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

  Further,
  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. Behind take any positive number. For example, let's find
  . Let
  . Then
  ,. Thus,
  .

  3)
  .

  We have
  . Because the
  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 the
  , we have
  .

  So,
  .

  4)
  , on 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,
  .

  Definition 1. A sequence is called non-decreasing [non-increasing] if each element of the sequence, starting from the second, is not less than [not more than] its previous element, that is, if the inequality is true for all numbers

  Definition 2. A sequence is called monotonic if it is either non-decreasing or non-increasing.

  If the elements of a non-decreasing sequence for all numbers satisfy a strict inequality, then this sequence is called increasing.

  Similarly, if the elements of a non-increasing sequence for all numbers satisfy a strict inequality, then this sequence is called decreasing.

  Note that every monotonic sequence is obviously bounded on one side (either from above or from below). Indeed, every non-decreasing sequence is bounded from below (the value of its first element can be taken as the lower bound), and every non-increasing sequence is bounded above (the value of its first element can also be taken as the upper bound).

  It follows that a non-decreasing sequence will be bounded on both sides, or simply bounded, if and only if it is bounded above, and a non-increasing sequence will be bounded if and only if it is bounded below.

  Let's look at examples of monotonic sequences.

  1. The sequence is non-decreasing. It is limited from below by the value of its first element, but is not limited from above.

  2. The sequence is decreasing. It is limited on both sides: from above by the value of its first element 2, and from below, for example, by the number 1.