Download §1. Basic definitions Let IR be the set of all real numbers, while IR

§1. Basic definitions
S
Let IR be the set of all real numbers, while IR := IR {−∞; ∞} be an extended set
of real numbers, and IN := {1, 2 . . . , n, . . .} be the set of all natural numbers.
Definition 1.1. Let X be a set of elements of arbitrary nature. For any n ∈ IN, by
xn we denote an element of X corresponding to n. This defines a set of numbered
elements {x1 , x2 , . . . , xn , . . .} (or, for brevity, {xn }) called a sequence defined on X.
If X = IR then {xn } is called a sequence of numbers. If X is a set of functions:
xn = xn (t), then {xn (t)} is a sequence of functions.
Definition 1.2. A sequence {xn } is called stationary if xn = xn+1 ∀n ∈ IN.
Definition 1.3. Two sequences {xn } and {yn } are called equal if xn = yn ∀n ∈ IN.
Sequences can also be defined as mappings from IN to X. More precisely,
Definition 1.4. A sequence is a single-valued mapping from IN to X.
In this study, we mainly consider sequences of numbers, and for simplicity refer
to them as to sequences.
Definition 1.5. The sum (difference, product, quotient) of two sequences {xn }
and {yn } is the sequence {zn } such that
zn = xn + yn ( xn − yn , xn · yn , xn /yn ) ∀ n ∈ IN,
where the quotient makes sense only if yn 6= 0 ∀ n ∈ IN.
Definition 1.6. A sequence {xn } is called bounded below (above) if ∃K ∈ IR
such that
xn ≥ K ( xn ≤ K ) ∀n ∈ IN.
Definition 1.7. A sequence that is bounded below and above is called bounded.
Or, equivalently
Definition 1.8. A sequence {xn } is called bounded if ∃K ∈ IR such that
|xn | ≤ K
∀n ∈ IN.
Counterdefinitions are extremely important for better understanding of new
concepts. For example, from Definition 1.8 we can define an unbounded sequence.
Definition 1.9. A sequence {xn } is called unbounded if ∀K ∈ IR ∃nK ∈ IN such
that
|xnK | > K.
Remark 1.10. Clearly, for any number K ∈ IR there exist infinitely many terms
of the sequence satisfying the inequality in Definition 1.9. If it were not true the
1
sequence would be bounded. Analogous statements hold for not bounded below
and not bounded above sequences.
Examples:
1. the sequence {n2 } is bounded below: n2 > 0 ∀n ∈ IN but not bounded above;
2. the sequence {−n} is bounded above: −n < 0 ∀n ∈ IN but not bounded below;
3. the sequence {(−1)n + 1} is bounded : |(−1)n + 1| ≤ 2 ∀n ∈ IN.
Infinitely large sequences represent an important subset of unbounded sequences.
Definition 1.11. A sequence {xn } is called infinitely large if ∀K ∈ IR ∃nK ∈ IN
such that
|xn | > K ∀n ≥ nK .
As an example, we show that the sequence {(−1)n n3 } is infinitely large.
Indeed, for any number K, we can find nk such that √
|(−1)n n3 | > K ∀n
√ ≥ nK .
To this end, we solve the inequality n3 > K, and n > 3 K. t Let nk = b 3 Kc + 1,
where bcc is the integer part of c. Then for n ≥ nK we obtain
√
3
n ≥ nk > K ⇒ n3 > K ⇒ |(−1)n n3 | > K.
From Definitions 1.9 and 1.11 it follows that any infinitely large sequence is
unbounded. However, the converse is not true: there exist unbounded sequences
that are not infinitely large. For example, such is the sequence {(1 − (−1)n )n}.
The concept of a limit plays a very important role in mathematics.
Definition 1.12. A number a ∈ IR is called a limit of a sequence {xn } if ∀ε > 0
∃nε ∈ IN such that
|xn − a| < ε ∀n ≥ nK .
Remark 1.13. For a convenience, we let the limit of an infinitely large sequence
{xn } be the symbol ∞:
lim xn = ∞.
n→∞
If {xn } is infinitely large and bounded below (above) then we write
lim xn = +∞, ( lim xn = −∞).
n→∞
n→∞
The statement ”{xn } converges to a” means that the limit of xn is a number
(or a symbol) a, and is denoted as {xn } → a as n → ∞. When a sequence has a
limit, it is called convergent. Otherwise, it is divergent.
We say that a sequence converges in IR, if either it has a finite limit or it
is infinitely large, i.e., if the sequence has a finite or infinite limit. A sequence
converges in IR if it has a finite limit, which may be denoted as
lim xn < ∞.
n→∞
2
Definition 1.14. A sequence {xn } is called infinitely small if
lim xn = 0,
n→∞
that is for any ε > 0 there exists nε such that
|xn | < ε ∀ n ≥ nε .
For example, the sequence {q n } for |q| < 1 is infinitely small. Indeed, for any
ε > 0 let us find nε such that |q n | < ε ∀ n ≥ nε . To this end, we solve the inequality
|q n | < ε, assuming that 0 < ε < 1 (for ε ≥ 1, the inequality is clearly true for any
n ∈ IN):
j ln ε k
ln ε
, nε =
+ 1,
n ln |q| < ln ε ⇒ n >
ln |q|
ln |q|
where ln ε < 0, and ln |q| < 0, since ε < 1, and |q| < 1. Thus, for n ≥ nε we have
n ≥ nε >
ln ε
< ln ε ⇒ |q|n < ε.
ln |q|
The fact that {xn } is not infinitely small means the following: there exists
ε0 > 0 such that for any n ∈ IN there exists kn > n with |xkn | > ε0 .
§2. Geometric interpretations of the main concepts
We will give geometric interpretations of the concepts introduced in Sect. 1 arising
from Definition 1.4, where sequence is defined as a function, see Fig. 1.
6
-
0 1 2 3 ...
n
n+1
Fig. 1. Sequence as a function.
From the geometric point of view, a sequence {xn } is bounded above (below)
if there exists a horizontal line y = K such that all the points (n, xn ) are below
(above) this line, see Fig. 2.
A sequence {xn } is bounded if there exist two horizontal lines y = K, and
y = −K such that all the points (n, xn ) are between these lines, see Fig. 3.
3
6
K
-
0 1 2 3 ...
n
n+1
Fig. 2. Bounded above sequence.
K
6
-
0 1 2 3 ...
n
n+1
-K
Fig. 3. Bounded sequence.
6
a+ε
a
a−ε
-
0 1 2 3 ...
nε
n
n+1
Fig. 4. Limit of a sequence.
A number a is a limit of the sequence {xn } if for any ε > 0, starting with some
nε all the points (n, xn ) are between two horizontal lines y = a − ε, and y = a + ε,
see the shaded area in Fig. 4. The points (n, xn ) tend to concentrate around the
line y = a.
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A sequence {xn } is infinitely small if for any ε > 0 starting with some nε all
the points (n, xn ) are between two horizontal lines y = −ε, and y = +ε. The points
(n, xn ) tend to concentrate around the line y = 0.
A sequence {xn } is infinitely large if for any ε > 0 only a finite number of
points (n, xn ) are between the two horizontal lines y = −ε, and y = +ε.
§3. Properties of convergent sequences
In this section we consider sequences convergent in IR (unless explicitly stated
otherwise).
Theorem 3.1. A convergent sequence has a unique limit.
Proof: Assume that a sequence {xn } has two limits: a and b (a 6= b). Then from
Definition 1.12 with ε = |b − a|/4, it follows that there exists a number n0 such that
|xn − a| < |b − a|/4, and |xn − b| < |b − a|/4 ∀n ≥ n0 .
Then
|b − a| = |(xn − a) + (b − xn )| ≤ |xn − a| + |xn − b| < |b − a|/2,
leading to a contradiction. The proof is complete.
Theorem 3.2. A convergent sequence is bounded.
Proof: Let {xn } → a. Then there exists n1 ∈ IN such that
|xn − a| < 1 ∀ n ≥ n1 .
Then
|xn | − |a| ≤ |xn − a| < 1 ⇒ |xn | < |a| + 1 ∀ n ≥ n1 ,
and
|xn | ≤ K = max(|x1 |, . . . , |xn1 −1 , |a| + 1) ∀ n ∈ IN.
The proof is complete.
Theorem 3.3. If lim xn = a then for any number b > a (b < a) there exists
n→∞
nb ∈ IN such that
xn < b (xn > b) ∀n ≥ nb .
Proof: Let b > a. From Definition 3.4 it follows that there exists nb such that
|xn − a| < b − a ∀n ≥ nb .
Rewriting this inequality yields
−(b − a) < xn − a < b − a ⇒ xn < b ∀n ≥ nb .
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The proof is complete.
Corollary 3.5. Let lim xn = a.
n→∞
1. If a 6= 0 then there exists n0 such that xn 6= 0 ∀n ≥ n0 .
2. If a > 0 (a < 0) then there exists n0 such that xn > 0 (xn < 0) ∀n ≥ n0 .
Theorem 3.6. Let lim xn = a.
n→∞
1. If there exists nb such that xn ≤ b (xn ≥ b) ∀n ≥ nb , then a ≤ b (a ≥ b).
2. If there exists nb such that xn < b (xn > b) ∀n ≥ nb , then a ≤ b (a ≥ b)
(where the strict inequality in general does not hold).
Proof: Assume xn ≤ b ∀n ≥ nb . Suppose a > b. Then from Theorem 3.3 there
exists nb such that xn > b ∀n ≥ nb . This inequality for n > max{nb , nb } leads to a
contradiction. The following example shows that in general the strict inequality in
part 2 does not hold.
1
> 0 = b ∀n ∈ IN;
n
1
= 0 = a; a = b = 0.
n→∞ n
lim
The proof is complete.
Theorem 3.7. If starting from some n0 , xn ≤ yn , and {xn }, {yn } are both
convergent, then
lim xn ≤ lim yn .
n→∞
n→∞
Proof: Let
lim xn = a,
n→∞
lim yn = b, c = (a + b)/2.
n→∞
If a > b, then a > c > b, and from Theorem 3.3 there exist n0c , n00c such that
xn > c ∀n ≥ n0c , yn < c ∀n ≥ n00c .
Then for all n ≥ max{n0c , n00c } both inequalities hold true, that is xn > c > yn ,
leading to a contradiction. The proof is complete.
Theorem 3.8. Let sequences {xn }, {yn }, and {zn } be such that there exists
n0 ∈ IN with the property that
either xn ≤ yn ≤ zn , or xn ≥ yn ≥ zn ∀n ≥ n0 .
If {xn } and {zn } converge to the same limit (finite or infinite), then {yn } converges
to the same limit as well.
Proof: Let {xn } → a, {zn } → a, and a ∈ IR. By the assumption of the theorem,
xn − a ≤ yn − a ≤ zn − a ∀n ≥ n0 .
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On the other hand, for any ε > 0 there exist n0ε , n00ε ∈ IN such that
|xn − a| < ε ⇔ −ε < xn − a < ε ∀n ≥ n0ε ,
|zn − a| < ε ⇔ −ε < zn − a < ε ∀n ≥ n00ε .
For n ≥ nε = max(n0 , n0ε , n00ε ) all the above inequalities hold true. Therefore,
−ε < xn − a ≤ yn − a ≤ zn − a < ε ∀n ≥ nε .
This implies that {yn } → a. For a = ±∞ the proof is very similar.
Sometimes, Theorem 3.8 is called the ”Squeeze” theorem.
Theorem 3.9.
1. If sequences {xn } and {yn } converge then their sum, difference, and product
converge as well. Moreover,
lim (xn ± yn ) = lim xn ± lim yn ,
n→∞
n→∞
n→∞
lim (xn · yn ) = lim xn · lim yn .
n→∞
n→∞
n→∞
2. Additionally, if yn 6= 0 ∀n ∈ IN, and lim yn 6= 0 then
n→∞
lim (xn /yn ) = lim xn / lim yn .
n→∞
n→∞
n→∞
Proof: Let {xn } → a, and {yn } → b. By Theorem 3.2, the sequence {yn } is
bounded:
∃K ∈ IR : |yn | ≤ K ∀n ∈ IN.
Then, for ε > 0 there exist n0ε and n00ε such that
|xn − a| <
ε
ε
∀n ≥ n0ε , |yn − b| <
∀n ≥ n00ε ,
2M
2M
where M = max{1, K, |a|}. For n ≥ nε = max{n0ε , n00ε } both inequalities hold true,
and
|(xn ± yn ) − (a ± b)| ≤ |xn − a| + |yn − b| < ε/M ≤ ε,
|xn yn − ab| = |(xn yn − yn a) + (yn a − ab)| ≤
Kε
|a|ε
+
≤ ε.
2M
2M
Therefore, {xn ± yn } → (a ± b), and {xn yn } → ab.
We now prove part 2. First we show that the sequence {1/yn } is bounded, and
{1/yn } → 1/b. Indeed, by Definition 3.4 there exists nb such that
|yn ||xn − a| + |akyn − b| <
|yn − b| < |b|/2 ∀n ≥ nb .
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Then for all n ≥ nb
1
2
|b| − |yn | ≤ |b − yn | < |b|/2 ⇒ |yn | > |b|/2 ⇒ <
.
yn
|b|
Next, for ε > 0 there exits nε ≥ nb such that
|yn − b| < εb2 /2 ∀n ≥ nε .
Thus, for n ≥ nε
1
−
yn
1 |yn − b|
|yn − b|
<
< ε,
=
b
|yn | · |b|
|b| · |b|/2
and { y1n } → 1b . Therefore, {xn /yn } → a · 1b . The proof is complete.
Corollary 3.10. The sum, difference and product of infinitely small sequences
are infinitely small as well.
Theorem 3.11. If {xn } is infinitely small, and {yn } is bounded then {xn · yn } is
infinitely small.
Proof: Let {xn } → 0, and |yn | ≤ K ∀n ∈ IN. Then for ε > 0 there exists nε such
that
|xn | < ε/K ∀n ≥ nε .
Then
|xn · yn | = |xn | · |yn | < ε/K · K = ε.
The proof is complete.
Theorem
3.12. If {xn } is infinitely large, and xn 6= 0 ∀n, then the reciprocal
1
is
infinitely
small.
xn
Proof: By Definition 1.11, for any positive ε there exists nε such that
|xn | >
Therefore,
Hence,
1
xn
1
xn
1
> 0 ∀n ≥ nε .
ε
is defined and
1 < ε ∀n ≥ nε .
xn
is infinitely small.
The converse statement is also true, and the proof is very similar to that of
Theorem 3.12.
Theorem 3.13.
If {xn } is infinitely small, and xn 6= 0 ∀n ∈ IN, then the
reciprocal x1n is infinitely large.
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§4. Monotonic sequences
Definition 4.1. A sequence {xn } is called non-decreasing (non-increasing) if
xn ≤ xn+1
(xn ≥ xn+1 ) ∀n ∈ IN.
A sequence {xn } is called strictly increasing (strictly decreasing) if
xn < xn+1
(xn > xn+1 ) ∀n ∈ IN.
Definition 4.2. A sequence {xn } is called monotonic if it is either non-decreasing
or non-increasing. A sequence {xn } is called strictly monotonic if it is either strictly
decreasing or strictly increasing.
Theorem 4.3. If {xn } is monotonic then it converges in IR. Moreover, if {xn } is
non-decreasing then
lim xn = sup{xn },
x→∞
if {xn } is non-increasing then
lim xn = inf{xn }.
x→∞
A monotonic sequence converges in IR if and only if it is bounded.
Proof: Let {xn } be non-decreasing, and sup{xn } = a < ∞. Then every xn ≤ a,
and for ε > 0 there exists xnε such that
xnε > a − ε.
On the other hand, xn ≥ xnε ∀n ≥ nε since {xn } is non-decreasing. Hence, for
n ≥ nε
a − ε < xnε ≤ xn ≤ a < a + ε ∀n ≥ nε .
Therefore, {xn } → a.
If {xn } is non-decreasing and unbounded, then it is bounded below, and
sup{xn } = +∞. Moreover, {xn } is infinitely large. Indeed, since it is not bounded
above, for any ε there exists xnε such that xnε > ε. Due to monotonicity, xn ≥
xnε > ε ∀n ≥ nε . Therefore, {xn } → +∞ (see Remark 1.13), and the proof is
complete.
Example. We will show that the sequence {(1 + 1/n)n } is strictly increasing,
bounded, and by Theorem 4.3 has a finite limit: we denote it by e:
lim (1 + 1/n)n =: e.
x→∞
Let xn = (1 + 1/n)n . We first show that xn < xn+1 ∀n ∈ IN. From binomial
expansion we obtain:
Another important application of Theorem 4.3 is given in the following lemma.
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Lemma 4.4 (On nested intervals). Let the sequence of the intervals {[an , bn ]}
be such that [an+1 , bn+1 ] ⊂ [an , bn ] ∀n ∈ IN (such intervals are called nested), and
their lengths converge to zero, i.e., lim (bn − an ) = 0. Then there exists a unique
n→∞
point ξ that belongs to every interval, i.e., ξ ∈ [an , bn ] ∀n ∈ IN, and
ξ = lim an = sup{an } = lim bn = inf{bn }.
n→∞
n→∞
Proof: The sequences {an } and {bn } are monotonic and bounded:
a1 ≤ an ≤ an+1 ≤ bn+1 ≤ bn ≤ b1
∀n ∈ IN.
By Theorem 3.4, they are convergent, and
lim an = sup{an } =: a,
n→∞
lim bn = inf{bn } =: b.
n→∞
However, lim (bn − an ) = 0 implies that a = b =: ξ, and clearly
n→∞
an ≤ a = ξ = b ≤ bn
∀n ∈ IN.
Next we show that there is no point other than ξ that belongs to all the intervals
[an , bn ]. Assume the opposite: ∃η ∈ [an , bn ] ∀n and η 6= ξ. Then by Theorem 3.6
bn − an ≥ |ξ − η| > 0 ∀n ⇒
lim (bn − an ≥ |ξ − η| > 0.
n→∞
This contradiction completes the proof of the lemma.
§5. Partial limits
So far we have been studying convergent sequences. This raises a natural question:
What about divergent ones? Can they be studied? What characteristics need to
be defined for that? It turns out that divergent sequences also can be describes and
thoroughly studied. In order to do that we first introduce the following important
concept.
Definition 5.1. Let xn be a sequence, and let nk be a strictly increasing sequence
of natural numbers. Then the sequence xnk is called a subsequence of xn .
Clearly, every sequence has an infinite number of subsequences. Next we prove
the following important theorem.
Theorem 5.2 (Bolzano-Weierstrass). Every bounded sequence contains a convergent in IR subsequence.
Proof: Let {xn } be bounded. Then there exists an interval [−K, K] containing
all terms of {xn }. Let a1 := −K, b1 := K. Let c1 = (a1 + b1 )/2. Then one of
the intervals, either [a1 , c1 ] or [c1 , b1 ], contains infinitely many terms of {xn } (both
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can not contain only finite number of the terms, because the whole sequence is
infinite). We denote that interval by [a2 , b2 ]. If both [a1 , c1 ] and [c1 , b1 ] contain
infinitely many terms, then by [a2 , b2 ] we denote either one.
Next, let c2 = (a2 +b2 )/2. By [a3 , b3 ] we denote that one of the intervals [a2 , c2 ]
and [c2 , b2 ] which contains infinitely many terms of {xn }. Again, if both contain
infinitely many terms, then we choose either one.
Continuing this process, let cn = (an + bn )/2. By [an+1 , bn+1 ] we denote that
one of the intervals [an , cn ] and [cn , bn ] which contains infinitely many terms of
{xn }. If both contain infinitely many terms, then we choose either one.
By this process, we obtain a sequence of nested intervals {[an , bn ]}: since
[an+1 , bn+1 ] ⊂ [an , bn ]. The lengths of the intervals go to zero:
4K
= 0.
n→∞ 2n
lim (bn − an ) = lim
n→∞
According to the lemma 4.4 on nested intervals there exists a point ξ such that
ξ ∈ [an , bn ] ∀n ∈ IN and
ξ = lim an = lim bn .
n→∞
n→∞
Since every interval [an , bn ] contains infinitely many terms of {xn }, for any k ∈ IN
there exists a term xnk ∈ [ak , bk ] with nk > nk−1 . By Theorem 1.6 the subsequence
{xnk } converges to ξ:
ak ≤ xnk ≤ bk ∀k, {ak } → ξ,
{bk } → ξ.
The proof is complete.
For unbounded sequences the following theorem holds.
Theorem 5.3. Every sequence that is not bounded above (below) contains a
subsequence that converges to +∞ (−∞). Moreover, if a sequence is unbounded
and not infinitely large, then it contains a subsequence convergent in IR.
Proof:
Combining Theorem 5.2 and Theorem 5.3 we conclude the next theorem.
Theorem 5.4 . Every sequence contains a convergent in IR. subsequence. Moreover, if a sequence is not infinitely large, then it contains a subsequence convergent
in IR.
Therefore, every sequence can be studied by analyzing its convergent subsequences. There may be many of them, and they can converge to different numbers
in IR. These numbers characterize the sequence.
Definition 5.5. A number a is called a partial limit of xn if there exists {xnk } –
a subsequence of {xn } – such that {xnk } → a.
Thus, every sequence can be characterized by the set of its partial limits. In
the following sections, we denote by Lim xn the set of partial limits of {xn }.
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§6. Upper and Lower limits
First we introduce the following important characteristics of sequences.
Definition 6.1. The quantities
sup Lim xn =: lim xn , and inf Lim xn =: lim xn
are called the upper and the lower limits of {xn } respectively.
Now we establish some properties of the upper and the lower limits. First of
all we note that every sequence has upper and lower limits, and they are unique.
If a sequence is not bounded above (below), then by Theorem 5.3
lim xn = +∞
(lim xn = −∞).
Clearly, for any sequence
lim xn ≥ lim xn .
From the properties of supremum and infimum it follows that
lim (−xn ) = −lim xn , lim (−xn ) = −lim xn .
Moreover, for any {xnk } subsequence of {xn }
lim xn ≥ lim xnk , lim xn ≤ lim xnk .
Theorem 6.2. Let {xn } be convergent. Then
Lim xn = { lim xn },
n→∞
and the following equalities hold
lim xn = lim xn = lim xn .
n→∞
Proof: We claim that if {xn } → a, then {xnk } → a for any subsequence {xnk } of
{xn }. Indeed, let lim xn = a. Then for any positive ε there exists nε such that
n→∞
|xn − a| < ε ∀n ≥ nε .
In particular,
|xnk − a| < ε ∀nk ≥ nε .
Therefore, {xnk } → a as well for any subsequence. This means that Lim xn = {a},
and the proof is complete.
Theorem 6.3. The upper and the lower limits are partial limits, i.e.,
lim xn ∈ Lim xn ,
lim xn ∈ Lim xn .
Proof:
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§7. Criteria of convergence
In this section we establish several important criteria of convergence for sequences.
The first one is trivial.
Theorem 7.1. A sequence is convergent (in IR) if and only if its every subsequence
is convergent (in IR).
Proof: (⇒) follows from Theorem 6.2. (⇐) follows from the fact that the sequence
itself is its own subsequence.
Theorem 7.2. A sequence is convergent (in IR) if and only if
lim xn = lim xn .
Their common value is the limit of the sequence.
Proof: (⇒) follows from Theorem 6.2. (⇐) follows from Theorem 7.1.
Theorem 7.3. A sequence is convergent (in IR) if and only if the set of its partial
limits consists of one point.
Proof: (⇒) follows from Theorem 6.2. (⇐) follows from Theorem 7.2.
To establish one more criterion of convergence, we need the following definition.
Definition 7.4. {xn } is called a fundamental (or Cauchy) sequence if ∀ε > 0 ∃nε
such that
|xn+m − xn | < ε ∀n ≥ nε ∀m ∈ IN.
Lemma 7.5. A fundamental sequence is bounded.
Proof: By Definition 7.4, ∃n1 such that
|xn+m − xn | < 1
∀n ≥ n1 ∀m ∈ IN.
In particular,
|xn1 +m − xn1 | < 1
∀m ∈ IN.
Then
|xn | ≤ K = max{|x1 |, . . . , |xn1 −1 |, |xn1 | + 1} ∀n ∈ IN.
The proof is complete.
Theorem 7.6 (Cauchy Criterion of Convergence). A sequence is convergent
(in IR) if and only if it is fundamental.
Proof: (⇒). Let lim xn = a < ∞. Then for any positive ε there exists nε such
n→∞
that
|xn − a| < ε/2 ∀n ≥ nε .
13
Therefore,
|xn+m − xn | = |(xn+m − a) + (a − xn )| ≤ |xn+m − a|+
|a − xn | < ε/2 + ε/2 = ε
∀n ≥ nε ,
∀m ∈ IN,
since n + m > n ≥ nε ∀m ∈ IN.
(⇐). Let {xn } be fundamental. By Definition 7.4, for any positive ε there
exists nε such that
|xn+m − xn | < ε/2 ∀n ≥ nε ∀m ∈ IN.
By Lemma 7.5, {xn } is bounded. By Bolzano-Weiestrass Theorem 5.2, {xn } contains a convergent subsequence {xnk }, i.e., lim xnk = a < ∞. we claim that
k→∞
lim xn = a. Indeed by Definition 3.4, for any positive ε there exists n0ε > nε such
n→∞
that
|xnk − a| < ε/2 ∀nk ≥ n0ε .
Combining these inequalities, for any n > n0ε we obtain:
|xn − a| = |(xn − xn0ε ) + (xn0ε − a)| ≤ |xn − xn0ε | + |xn0ε − a| < ε.
Thus, {xn } → a, and the proof is complete.
§8. Structures of sequences
In this section we summarize the theory of sequences, and consider structures of
sequences. The set of partial limits Lim xn is a very important characteristics of a
sequence.
First we characterize bounded sequences.
1. {xn } is bounded if and only if its upper and lowere limits are finite:
−∞ < lim xn ≤ lim xn < +∞.
2. ∀ε > 0 all terms of a bounded sequence with the exception of a finitely many
of them are located in the interval:
(lim xn − ε, lim xn + ε),
or, equivalently, all the points (n, xn ) with the exception of a finitely many of
them are located between the lines
y = lim xn − ε, and y = lim xn + ε.
3. ∀ε > 0 there are infinitely many terms of xn located above the line y =
lim xn − ε and below the line y = lim xn + ε.
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For unbounded sequences the following is true:
4. If {xn } is infinitely large, then its set of partial limits consists of not more that
two points: Lim xn ⊆ {−∞, +∞}. More precisely, if {xn } is bounded above
(below), then Lim xn = {−∞} (Lim xn = {+∞}. If it is neither bounded
above nor below, then Lim xn = {−∞, +∞}
Example: We characterize the sequence {xn } with Lim xn = {−∞, 0, 2}.
For this sequence we conclude that lim xn = −∞, lim xn = 2. It is divergent, not
bounded below, bounded above, not infinitely large. It contains an infinitely large
and infinitely small subsequences, as well as a subsequence convergent to 2.
§9. Problems
1. Characterize {xn } if
(a) Lim xn = {0, 2},
(b) Lim xn = {−∞, −1, 0, 2},
(c) Lim xn = {−1, 1, +∞},
(d) Lim xn = {−∞, 2, +∞},
(e) Lim xn = {0, 2, +∞}.
2. Characterize {xn } and { x1n } if
(a) Lim xn = {2},
(b) Lim xn = {−1},
(c) Lim xn = {−1, 1},
(d) Lim xn = {−∞, −2, 2},
(e) Lim xn = {2, 3, +∞}.
3. Characterize {xn · yn } if
(a) Lim xn = {0}, Lim yn = {−1, 1};
(b) Lim xn = {2}, Lim yn = {−2, 3};
(c) Lim xn = {−1, 2}, Lim yn = {2, 3}.
4. Find Lim xn for
(a) {(−1)n },
(b) {−n},
(c) {1 − (−1)n },
(d) {sin πn/3}.
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§10. Power Series
Definition 10.1. Let {an } be a sequence of real numbers. The series
∞
X
a n xn = a 0 + a 1 x + a 2 x2 + a 3 x3 + . . .
(10.1)
n=0
is called a power series.
an . Then (10.1) converges absolutely
Ratio Criterion 10.2. Let R := lim n→∞ an+1
whenever |x| < R and diverges whenever |x| > R.
Proof: Follows from the ratio test applied to (10.1).
e := lim |an |1/n . Then (10.1) converges absolutely
Root Criterion 10.3. Let R
n→∞
e and diverges whenever |x| > R.
e
whenever |x| < R
Proof: Follows from the root test applied to (10.1).
e This quantity is
From theorems 10.2 and 10.3 we comclude that R = R.
reffered to as the radius of convergence.
Theorem 10.4. Let R be the radius of convergence of (10.1). Then (10.1) converges uniformly on [−K, K] for any 0 < K < R.
Proof: Follows from the Weierstrass M-test.
Theorem 10.5. Let R be the radius of convergence of (10.1). If (10.1) converges
at x = R, then it converges uniformly on [−K, R] for any 0 < K < R. If (10.1)
converges at x = −R, then it converges uniformly on [−R, K] for any 0 < K < R.
HW Example. Consider
∞
X
xn /n!. Its radius of convergence is +∞. There-
n=0
fore, it uniformly converges on any closed interval [−K, K]. However, it does NOT
uniformly convergent on IR. Prove all the statements made above.
§11. Taylor Series
Theorem 11.1. Let f have n + 1 derivatives on (a, b), and its first n derivatives
be continuous on [a, b]. Then for each x ∈ [a, b] with x 6= x0 there exists c between
x and x0 such that
f (x) = f (x0 ) + f 0 (x0 )(x − x0 ) +
+
f (n) (x0 )
f 00 (x0 )
(x − x0 )2 + . . . +
(x − x0 )n +
2!
n!
f (n+1) (c)
(x − x0 )n+1 .
(n + 1)!
Proof: Fix x ∈ [a, b] with x 6= x0 , and let M be the unique solution of
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f (x) = f (x0 ) + f 0 (x0 )(x − x0 ) +
f (n) (x0 )
f 00 (x0 )
(x − x0 )2 + . . . +
(x − x0 )n +
2!
n!
+M (x − x0 )n+1 .
Our proof will consist of showing that M =
Define
f (n+1) (c)
(n+1)!
for some c between x and x0 .
f (n) (t)
(x − t)n + M (x − t)n+1 .
n!
Since F (x) = f (x), and by our choice of M , F (x0 ) = f (x), from the Mean Value
Theorem it follows that ∃ c between x and x0 such that
F (t) = f (t) + f 0 (t)(x − t) + . . . +
F (x) − F (x0 )
= 0.
x − x0
After computing the derivative of F , we conclude that
F 0 (c) =
0 = F 0 (c) =
f (n+1) (c)
(x − c)n − M (n + 1)(x − c)n ,
n!
and the result follows.
Definition 11.2. Let f ∈ C ∞ . The series
∞
X
f (n) (x0 )
(x − x0 )n
n!
n=0
(11.1)
is called the Taylor’s series about x0 generated by f . To indicate that f generates
this series, we write
∞
X
f (n) (x0 )
f (x) ∼
(x − x0 )n .
n!
n=0
Does (11.1) converge for any x besides x = x0 ? If so, is its sum equal to f (x)?
In general, the answer to both questions is ”No”. From Theorem 11.2 it follows
that a necessary and sufficient condition for the Taylor’s series to converge to f (x)
is that
f (n+1) (c)
lim
(x − x0 )n+1 = 0.
n→∞ (n + 1)!
In practice it may be quite dificult to deal with this limit because of the unknown
position of c. In some cases, however, the limit can be shown to be zero.
Theorem 11.3. Let f ∈ C ∞ on [a, b], and |f i (x)| ≤ M ∀ x ∈ [a, b] ∀i ∈ IN. Then
∀ x ∈ [a, b] we have
∞
X
f (n) (x0 )
f (x) =
(x − x0 )n .
n!
n=0
Proof: HW
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§12. Problems
For each of the following functions
(a) Write the Taylor’s series about x0 = 0;
(b) Find its radius of convergence;
(c) Determine whether the Taylor’s series converges to the generating function.
1. f (x) =
1
(1−x)
2. f (x) = ln(1 − x)
3. f (x) = ex
4. f (x) = sin x
5. f (x) = cos x
6. f (x) =
1
(1−x)2
7. f (x) =
√
1+x
8. f (x) = (1 + x)α , α ∈ IR
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