Why is the criterion for the convergence of a sequence called internal? Fundamental Sequences

CAUCHY CRITERION

1) K.K. convergence of a number sequence: in order for numbers (real or complex) xn,n=1, 2, . . ., had a limit, it is necessary and sufficient that for anyone there exists a number N such that for all carried out

The criterion for the convergence of a number sequence is generalized into a criterion for the convergence of points of a complete metric. space.

Sequence of points (x p) full metric space converges if and only if for any there exists such N, that the inequality holds for everyone

2) K.K. existence limit of functions of n variables Let f be defined on a set of Xre-dimensional space Rn and takes numeric (real or complex) values, A - limit point of a set X (or symbol, in this case X is unbounded). A finite limit exists if and only if for everyone there is such U=U(a) . points A, that for any and the inequality holds

This criterion generalizes to more general mappings: let X- topological , A - its limit point at which countability holds, Y- full metric space and f - Xв Y. For there to be a limit

it is necessary and sufficient for there to be a neighborhood for everyone U=U(a).points attacking that the inequality holds for all

3) Q. for uniform convergence of a family of functions. Let X- some set, Y- topological a space that satisfies the first axiom of countability at the limit point, R is a complete metric. space, f( x, y). - mapping of the set Family of mappings f( x, y), mapping for a fixed set X into H, is uniformly convergent on X if for any there exists such a neighborhood U=U(y 0).points y 0 that's for everyone and all inequality is satisfied

In particular, if Y- set of natural numbers and then the sequence converges uniformly on the set X if and only if for any there exists such a number N, that for all and all numbers the inequality holds

4)K. to the convergence of a series: numerical converges if and only if for any there is such a number N, that for any and all integers the inequality holds

For multiple series, a similar convergence criterion is called. Cauchy-Stolz criterion. For example, in order to

converged on rectangular partial sums

it is necessary and sufficient for anyone to find something like this N, that with everyone and everyone whole the inequality was satisfied

These criteria are generalized to series in Banach spaces (the norms of the corresponding elements are taken instead of the absolute value).

5) Q. for uniform convergence of a series: let be functions defined on a certain set X and taking numerical values. In order for the series

converged uniformly on the set X, it is necessary and sufficient that such a number exists for everyone N, that for all whole the inequality was satisfied

This criterion also extends to multiple series, not only to numerical series, but also to series whose terms belong to Banach spaces, i.e. when and p(x).are mappings of the set X into a certain swarm.

6) Q. for the convergence of improper integrals: let a function f be defined on a half-interval, take numerical values on it and be integrable for any value (Riemann or Lebesgue) on the interval [ a, c]. In order to

converged, it is necessary and sufficient that for anyone there exists such that for all satisfying the condition the inequality holds

The criterion is formulated in a similar way for improper integrals of other types, and is also generalized to the case when the function f depends on several variables and its values lie in a Banach space.

7) K.K. for uniform convergence of improper integrals: let the function f( x, y).for each fixed where Y- some set defined on a half-interval takes numerical values and is integrable over any interval [ a, c]. In order to

converges uniformly on the set Y, it is necessary and sufficient that for any there is such that for any satisfying the conditions and all the inequality holds

This criterion also extends to improper integrals of other types, to the case of functions of several variables and to functions whose values lie in Banach spaces.

Lit.: C a u c h u A. L., Analyze algebrique, P., 1821; Stolz O., "Math. Ann.", 1884, Bd 24, S. 154-71; Dieudonne J., Fundamentals of modern analysis, trans. from English, M., 1964; Il'in V.A., Poznya to E.G., Fundamentals of Mathematical Analysis, 3rd ed., vol. 1, M., 1971, vol. 2, M., 1973; Kudryavtsev L. D., Course of mathematical analysis, t. . 1 - 2, M., 1981; 16] Nikolsky S.M., Course of mathematical analysis, 2nd ed., vol. 1-2, M., 1975; Whittaker E. - T., V a tson J. - N., Course of modern analysis, trans. from English, 2nd ed., part 1, M., 1963. L. D. Kudryavtsev.

Mathematical encyclopedia. - M.: Soviet Encyclopedia. I. M. Vinogradov. 1977-1985.

See what the "CACHY CRITERION" is in other dictionaries:

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The Cauchy criterion for the convergence of a sequence implies the most general criterion for the convergence of a number series. Theorem 4 (Cauchy criterion). In order for the number series Y1 an to converge, it is necessary and sufficient that for any number e > O there is a number N = N(e) such that for any n > N the inequality holds for all Using the partial sums 5P+P and Sn-\ considered series J2 in> inequality (1) can be written in the form From the Cauchy criterion follows the necessary criterion for the convergence of a number series. Theorem 5. If the series Comparison test for series with positive terms D'Alembert's test Cauchy's test Cauchy's criterion for the convergence of a series converges, then Assuming in Theorem 4, we obtain an inequality that holds for all. Due to the arbitrariness of the number e > 0, this means that Corollary. If lim an is different from zero or does not exist, then the series Example 1. Number series diverges, since Example 2. The series diverges, since it does not exist. Comment. Theorem 5 gives a necessary condition for the convergence of a series, but it is not sufficient, i.e. the condition lim o„ = 0 can also be satisfied for a divergent series. Example 3. Consider a numerical series called a harmonic series. For the harmonic series, the necessary condition for convergence is satisfied, since Using the Cauchy criterion, we show that this series diverges. Let's put p-p. Then the resulting inequality is satisfied for any arbitrarily large n. It follows that for e ^ 5 and p = n inequality (1) does not hold. Thus, due to the Cauchy criterion, the harmonic series diverges. Important note. In a certain sense, a series is a generalization of a finite sum. However, unlike the latter, the terms in which can be grouped and rearranged completely arbitrarily, which is why the sum, as we know, does not change, actions with members of an arbitrary series must be carried out carefully - the consequences may not always be predictable. If in a divergent series (the necessary criterion for convergence is not fulfilled) we group neighboring groups in pairs, then we get a convergent series. The terms of a convergent series (see example from § 8) can be rearranged so that it converges to any number and even diverges. In particular, the series obtained by rearranging its terms converges to half the sum of the original one (example from § 9). The fact that in these examples the terms of the series have different signs is significant. Let us present signs that make it possible to establish the convergence or divergence of some numerical series by comparing them with other series, the convergence or divergence of which is known in advance. Theorem 6 (comparison test). Let two series be given whose terms an and 6„ are positive. If the inequality holds for all numbers n, then from the convergence of the series Y1 6n follows the convergence of the series an, and from the divergence of the series Y1 On follows the divergence of the series Y1 6„. M Let's compose partial sums of series (1) and (2) From condition (3) of the theorem it follows that 5П ^ Sn for all 1) Assume that series (2) converges, i.e. there is a limit of its nth partial sums So since all terms of these series are positive, then, due to inequality (3), it follows that Thus, all partial sums 5P of series (1) are limited and increase as n increases, since. Consequently, the sequence of partial sums is convergent, which means the convergence of the series an. In this case, when passing to the limit in the inequality, we obtain that By virtue of the inequality we obtain Comparison test for series with positive terms D'Alembert's test Cauchy's test Cauchy's criterion for the convergence of the series i.e. series bn diverges. Comment. Theorem 6 remains valid in the case when the inequality an ^ bn is satisfied not for all n, but only starting from a certain number A:, that is, for all n ^ Jfc, since changing the finite number of terms of the series does not violate its convergence. Examples. Examine the following series for convergence: We have Since the number series converges, then by comparison the original series (4) also converges. The inequality implies the inequality Since the harmonic series diverges (like the series, then by comparison the original series (4) also diverges. I Theorem 6 remains valid in the case of a more general inequality Example 3. Examine series 4 for convergence Using the inequality sin x ^ x, which is valid for all, we find that for. Since the series converges, then by comparison (here A = y) this series (5) also converges. Corollary: If there is a finite nonzero limit, then series (1) and (2) converge or diverge simultaneously. From the existence of the above limit it follows that for any number e > O, there is a number N. such that for all n > N the inequality or Hence If the series (2) converges, then the series converges. But since then, by virtue of Theorem 6, the series (1) will also converge. If the series (2) diverges, then it diverges. and the series (e is considered so small that. Since n is for everyone, according to Theorem 6, series (1) diverges. Comment. The condition of the lemma is equivalent to the fact that the sequences сс, and Lbn at are equivalent or, which is the same. In the case I = 0, the convergence of series (2) implies the convergence of series (1). The reverse is not true. In the case L = +oo, the divergence of series (1) implies the divergence of series (2). The reverse is not true. Examples. Examine the following number series for convergence: 4 Let us compare this series with the harmonic series. We have Since the harmonic series diverges, this series also diverges. Then the original series converges. §5. D'Alembert's test oo Theorem 7 (D'Alembert's test). Let a series an be given, where all an > 0. If there exists n =\ limit, then the series converges, and the series diverges.4 Let there exist a limit where Take q such that. Then for any number, for example, for e = , there is a number N such that for all n ^ N the inequality will hold. In particular, we will have from where for all From this inequality, giving n successively the values N, we obtain The terms of the series do not exceed the corresponding terms of the series which converges as a series composed of terms of a geometric progression with a denominator. By comparison, the series converges, which means that the original series an also converges. In the case starting from a certain number N, the inequality will be satisfied, or Consequently, an diverges, since the necessary - a sign of convergence. Comment. If or does not exist, then D'Alembert's test does not give an answer about the convergence or divergence of the series. Examples. Examine the following series for convergence: For a given series we have Comparison test for series with positive terms D'Alembert's test Cauchy's test Cauchy criterion for the convergence of a series By D'Alembert's test the series converges. We have This series diverges. . Cauchy test Theorem 8 (Cauchy test). Let the series oo be given. Take a number q such that. Since there is a limit where, then, starting from a certain number N, the inequality will hold. In fact, from the limit equality it follows that for any c, including for, there is a number N, starting from which the inequality whence A or, what is the same, From here we obtain for. Thus, all terms of the series, starting, are less than the corresponding terms oo of the convergent series £ 0π. By comparison, the series converges, and therefore series (1) also converges. Let it be. Then, starting from a certain number N, for all n > N, inequality > 1 will hold, or Consequently, series (1) diverges. Comment. If A = 1, then series (I) can either converge or diverge. Examples. Examine the following series for convergence: L We have The series converges. ^ m Here the series diverges. ^

Here we propose to consider a general sign of the existence of a finite limit for the sequence,
.

Definition 3.5. Subsequence ,
, is called fundamental if for an arbitrary number
there is such a number that's for everyone
inequality holds
.

The definition of a fundamental sequence is often conveniently used in the following form.

Definition 3.6. Subsequence is fundamental if for an arbitrary number
there is such a number that's for everyone
and any natural number inequality holds
.

Theorem 3.13 (Cauchy criterion). In order for a sequence to converge, it is necessary and sufficient that it be fundamental.

Proof. Necessity. Let the sequence ,
, converges, that is, exists
. Let's choose
. Then there is such a number that's for everyone
inequality holds:
.

Let
And
, Then

=

,

which means the sequence is fundamental.

Adequacy. Let the sequence is fundamental. Let us prove that it converges. The difficulty lies in finding such a number A, which is its limit.

Let's break the argument down into several steps.

a) Let us prove that the fundamental nature of the sequence implies its boundedness. Let's consider ε =1, then there is such a number n 1 that in front of everyone

n, m ≥ n 1 inequality holds
. In front of everyone n≥ n 1 fair:

Let , a, then for each natural the inequalities are satisfied
, that is limited.

b) Let's choose natural n. Consider the set
- a set of values of sequence members whose numbers are not less than the selected one n. By what was proved in a) the set X 1 limited. And from the obvious investments
it follows that each of these sets is bounded.

c) Consider two new sequences. To this end, for each set
let's denote:
,
. From the embeddings given in b) it follows that the sequence increases (
), and the sequence decreases (
). That's why
, that is, the sequences are monotonic and bounded and therefore converge. Note also that for all natural n inequalities are obvious
.

d) Let us prove that the difference of these two sequences tends to zero:
. Let us use the condition of fundamentality. For an arbitrary number
there is such a number that's for everyone k ≥ n ε the inequalities are satisfied
. These inequalities allow us to conclude that

at n≥ n ε . Hence,
.

e) By what was proved in part c) the sequence converges, let
. Because
and, then from the inequalities
and from the lemma about two policemen it follows that
. Sufficiency has been proven. The theorem has been proven.

3.9. Subsequences. Partial limits

Definition 3.7. Let ,
, is some numerical sequence and let ,
is a strictly increasing sequence of natural numbers. Then a sequence of the form
,
, is called a subsequence of the sequence .

If a sequence has no limit, then this does not exclude the possibility of the existence of a limit for some subsequence.

Definition 3.8. A partial limit of a sequence is the limit of some convergent subsequence.

Example 3.18. Let
. This sequence diverges (see Section 3.2), but its subsequences
And
converge to 1 and -1, respectively. So these numbers are partial limits of the sequence
.

Theorem 3.14. Let the sequence ,
, converges to the number a. Then any subsequence of it also converges to a.

Proof. Let
,
, - subsequence of the sequence ,
. Because is a strictly increasing sequence of natural numbers, then
in front of everyone
(this is easy to prove by induction). Let's choose
. By definition of convergence To a for everyone
the inequality will be satisfied
.The theorem has been proven.

Problem 3.14 Prove that for a sequence to converge it is necessary and sufficient that each of its subsequences converge.

Problem 3.15. Prove that from the conditions
a And
a it follows that
a.

Problem 3.16. Give an example of a sequence that has exactly ten partial limits.

Problem 3.17. Give an example of a sequence for which every real number is a partial limit.

Let us consider the question of the existence of partial limits in the case of a bounded sequence.

Theorem 3.15 (Bolzano-Weierstrass). Every bounded sequence contains a convergent subsequence.

Proof. Due to the limited nature of the sequence, we can specify the following numbers
that for anyone the inequalities are satisfied
. Divide the segment
in half. Then at least one half will contain an infinite number of terms of the sequence. This follows from the fact that the sequence consists of an infinite number of terms, and there are only two halves. Let us choose this half and denote it by
, if both are like that, then any of them.

Next, a segment
Let's divide in half again and choose the half containing an infinite number of terms of the sequence. Let us denote it by
. Continuing this process, -th step we get the segment
, which contains infinitely many terms of this sequence. Each of the constructed segments is contained in the previous one. Section length
equal to , that is, tends to zero with increasing . Applying Cantor's lemma on nested segments, we obtain that the sequences
And
tend to the general limit, we denote it by A.

Let us now construct a convergent to A subsequence. As choose any member of the sequence
contained in
. As
choose such a member of the sequence
, which is contained in
and number which is more (here it is used that the segment
contains infinitely many terms of the sequence). Arguing similarly, on -th step as
choose such a member of the sequence
, which is contained in
and number which is more
. Let us recall that each of the constructed segments contains infinitely many terms of the sequence, which determines the possibility of such a choice. Because
, A
, then by the lemma about two policemen
.The theorem has been proven.

We denote the set of all partial limits of a sequence by
. The proven Bolzano-Weierstrass theorem can be reformulated as follows:

every bounded sequence has a set
partial limits are not empty.

Additionally, we note that from the boundedness of the sequence, by the theorem on passage to the limit in inequalities, it follows that the set is bounded
. So there are many
has precise top and bottom edges.

Definition 3.9. Let ,
, is a bounded sequence, and let
is the set of all its partial limits. Values

,

are called the lower and upper limits of the sequence, respectively .

It does not directly follow from this definition that numbers ,belong to many
, but nevertheless fair

Theorem 3.16. The upper and lower limits of a limited sequence are its partial limits.

Proof. Let us show that there is such a subsequence
, What
. Because
<, then by definition of the exact upper bound there is from
, for which
. Next, there is

, for which
, and in general, for anyone there will be

, satisfying the inequalities:

Since every is a partial limit, then any neighborhood contains infinitely many sequence terms . Therefore there is a number , for which
; there is a number , for which

And
.

Continuing the reasoning, for everyone consider , satisfying the conditions

And
.

The subsequence constructed in this way
satisfies the inequalities

and by the lemma about two policemen tends to .

Similarly, a subsequence is constructed that converges to .The theorem has been proven.

From the proven theorem, in particular, it follows that there is no sequence such that the set of all partial limits of which is a bounded interval.

We will denote the upper and lower limits of the sequence by
And
respectively. As one of the characteristic properties of these quantities, we prove the following theorem.

Theorem 3.17 . Let – limited sequence,
;
. Then for any positive number each of the inequalities
And
satisfies only a finite set of terms of the sequence.

Proof. Let's assume the opposite. Let the set of numbers members of the sequence satisfying the inequality
, endlessly. Let's arrange these numbers in strict ascending order:
Then the subsequence
satisfies the inequalities
. According to the Bolzano-Weierstrass theorem, one can isolate from it a convergent subsequence, the limit which is more than . It's clear that

, and this contradicts the fact that - top edge. The resulting contradiction proves the theorem.

Definition. The sequence (x n) is called fundamental (Cauchy sequence), if for any e > 0 there is a number N such that for all numbers n, satisfying the condition n>=N, and for any natural number p(p=1,2,3...) the inequality is true:

|x n + p – x n |< e.

Theorem. (Cauchy criterion) . In order for the sequence (x n) to be convergent, it is necessary and sufficient that it be fundamental.

Proof.

1) Necessity. Let x n à a. We fix an arbitrary e > 0. Since the sequence (x n ) converges to the limit A, then for a number equal to e/2 there is a number N such that in front of everyone n >= N:

|x n – a|< e/2. (1)

If p any natural number, then for all n>=N it will be:

|x n + p – a| < e/2. (2)

Since the modulus of the sum of two numbers does not exceed the sum of their moduli, then from inequalities (1) and (2) we obtain for all n >= N and for any natural number p we will get:

|x n + p – x n | = | + |<= |x n + p – a| + |x n – a|< e, Þ |x n + p – x n | < e - this means that this is a fundamental sequence.

2) Adequacy. Let now (x n ) be a fundamental sequence. For example, for e =1 there is n 1 such that n > n 1 and m > n 1 has |x n - x m |< 1.

Fixing m o > n 1 we have |x n - x m o |< 1 и Þ |x n | < 1+ |xm o |

Þ |x n |<= M, где M=max{|x1|,…|xn1|,1+|xm o |) for all nÎN, i.e. (x n) – limited.

This means that by the Bolzano-Weierstrass theorem there exists a convergent sequence ( x n k ), x n k –> a. Let us show that (x n ) converges to a.

For a given e > 0:

"e > 0 $K(e)О N:"k>K(e) Þ

|x n k – a| < e;

In addition, due to the fundamental nature of (x n), $n e = n(e): n k ,n > n e

Þ |x n – x n k |< e/2

Let's put n e = max(n e , n k (e) ) and fix n ko > n e. then for n > n e we have:

|x n – a|<= |x n – xn ko | + |x n ko – a|< e. А это и означает, что lim x n = a #

15. Two definitions of the limit of a function at a point and their equivalence.

Def.1. (according to Cauchy). Let the function y=f(x) be given: X à Y and a point a is the limit for the set X. The number A called limit of the function y=f(x) at the pointa , if for any e > 0 it is possible to specify d > 0 such that for all xÎX satisfying the inequalities 0< |x-a| < d, выполняется |f(x) – A| < e.

Def.2. (according to Heine). Number A is called the limit of the function y=f(x) at the point a, if for any sequence (x n )Ì X, x n ¹a "nОN, converging to a, the sequence of function values (f(x n)) converges to the number A.

Theorem. Determination of the limit of a function according to Cauchy and according to Heine are equivalent.

Proof. Let A=lim f(x) be the limit of the function y=f(x) according to Cauchy

and (x n )Ì X, x n ¹a "nОN – sequence converging to a, x n à a.

Given e > 0, we find d > 0 such that at 0< |x-a| < d, xÎX имеем |f(x) – A| < e,

and from this d we find a number n d =n(d) such that for n>n d we have 0< |x n -a| < d.

But then |f(x n) – A| < e, т.е. доказано, что f(x n)à A.

Let now the number A there is now a limit of the function according to Heine, but A is not a Cauchy limit. Then there is e o > 0 such that for all nОN there exist x n ОX,

0 < |x n -a| < 1/n, для которых |f(x n)-A| >= e o . This means that the sequence (x n )Ì X, x n ¹a "nОN, x n à has been found a such that

sequence (f(x n)) does not converge to A. #

Uniqueness of the limit of a function at a point. Local boundedness of a function that has a finite limit. Local preservation of the sign of a function that has a zero limit.

Theorem 1. If $ lim f(x) = b О R for x à a, then this limit the only one.

Proof: Let it not be so.

lim f(x) = b 1 and lim f(x) = b 2 for x à a. b 1 ¹b 2

"(x n )О D(f), x n à a, x n ¹ a Þ f(x n) à b 1 (definition according to Heine)

"(x n )О D(f), x n à a, x n ¹ a Þ f(x n) à b 2 (definition according to Heine)

For a specific sequence (x n )М D(f). x n ’ à a, x n ¹ a Þ

Þ f(x n ’) à b 1 and f(x n ’)à b 2. Then, by the theorem on the uniqueness of the limit of the sequence, b 1 =b 2. #

Def. A function f(x) is said to be locally bounded for x à a if there are numbers d > 0 and M > 0 such that for 0< |x-a| < d, xÎX имеем |f(x)|<=M.

Theorem 1 (about local boundedness). If a function f(x) has a limit at a point a, then it is locally bounded for x à a.

Proof: If there exists lim f(x) = A for x à a, then, for example, for e=1 there exists d>0 such that for 0< |x-a| < d, xÎX, имеем |f(x)-A| < 1, а это значит,

|f(x)|<|A|+1=M. #

Theorem 2 (on local sign conservation). If lim f(x) = A for x à a and A¹0, then there exists d>0 such that for

0 < |x-a| < d, xÎX и A>0 we have f(x)>A/2, and at 0< |x-a| < d, xÎX и A<0 имеем

f(x)< a/2, т.е. (0 < |x-a| < d)L(xÎX) Þ |f(x)| >|A|/2.

Proof: Let's take e=|A|/2. There is d>0 such that for

0 < |x-a| < d, xÎX имеем

A-|A|/2

For A>0, from the left inequality we obtain f(x) > A/2, and for A<0 из правого неравенства получаем f(x) < A/2. #

Subsequence (xn) satisfies Cauchy condition, if for any positive real number ε > 0 there is a natural number N ε such that
(1) |x n - x m |< ε при n >N ε , m > N ε .

Sequences satisfying the Cauchy condition are also called fundamental sequences.

The Cauchy condition can be presented in another form. Let m > n. If m< n , то поменяем n и m местами. Случай нас не интересует, поскольку при этом неравенство (1) выполняется автоматически. Имеем:
;
.
Here p is a natural number.

Then the Cauchy condition can be formulated as follows:

Consistency satisfies Cauchy condition, if for any there is a natural number such that
(2) for and any natural p .

The number appearing in the Cauchy condition depends on ε. That is, it is a function of a real variable ε, the range of which is the set of natural numbers. The number can also be written in the form , as is customary for denoting functions.

Cauchy criterion for sequence convergence

In order for a sequence to have a finite limit, it is necessary and sufficient that it satisfies the Cauchy condition.

Proof of the Cauchy criterion for the convergence of a sequence

Proof of Necessity

Let the sequence converge to a finite limit a:
.
This means that there is some function so that for any the following inequalities hold:
(1.1) at .
See Definition of sequence limit.

Let us show that the sequence satisfies . To do this, we need to find a function such that, for any , the following inequalities are satisfied:
at .
Let us use the properties of inequalities and apply (1.1):
.
The last inequality holds for .

Let's replace it with . Then for any we have:
at ,
Where .

The need has been proven.

Proof of sufficiency

Let the sequence satisfy . Let us prove that it converges to a finite number. We divide the proof into three parts. First we prove that the sequence is bounded. Then we apply , according to which a bounded sequence has a subsequence that converges to a finite number. And finally, we will show that the entire sequence converges to this number.

Let us prove that the sequence satisfying is bounded. To do this, in the Cauchy condition, we set . Then there is a natural number for which the following inequalities hold:
(2.1.1) at .

Let's take any natural number and fix a member of the sequence. Let's denote it as to emphasize that this is a constant number that does not depend on the index n.

We substitute in (2.1.1) and perform transformations. When we have:
;
;
;
;
.
This shows that for , the terms of the sequence are limited. Since, for , there is only a finite number of terms, then the entire sequence is limited.

Let us apply the Bolzano–Weierstrass theorem. According to this theorem, a bounded sequence has a subsequence that converges to some finite number a. Let us denote such a subsequence as . Then
.

Let us show that the whole sequence converges to the number a.
Since the sequence satisfies , there is some function for which the following inequalities hold for any:
at .
Let us take the term of the convergent subsequence as the term and replace ε 1 by ε /2 :
(2.3.1) at .

Let us fix n. Then (2.3.1) is an inequality containing a sequence in which a finite number of first terms with are excluded. A finite number of first terms does not affect the convergence (see Influence of a finite number of terms on the convergence of a sequence). Therefore, the limit for a truncated sequence is still a. Applying properties of limits associated with inequalities And arithmetic properties of limits, for , from (2.3.1) we have:
at .
Let's use the obvious inequality: . Then
at .

That is, for any there is a natural number, so
at .
This means that the number a is the limit of the entire sequence (and not just its subsequence.

The theorem is proven

Used literature:
O.V. Besov. Lectures on mathematical analysis. Part 1. Moscow, 2004.