A proof of Weierstrass's limit theorem is given monotonic sequence. The cases of bounded and unbounded sequences are considered. An example is considered in which it is necessary, using Weierstrass’s theorem, to prove sequence convergence and find its limit.
Any monotone bounded sequence (xn) It has final limit, equal to the exact upper boundary, sup(xn) for non-decreasing and exact lower limit, inf(xn) for a non-increasing sequence.
Any monotonous unlimited sequence It 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
.
It 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 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 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 such 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 has been 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's prove that given sequence limited 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.
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 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 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 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 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, 
.