The Algebra of the Reals
In this article, we introduce key mathematical notations and symbols, beginning with the set of natural numbers, denoted as N = {1, 2, 3, }, and the integers represented by Z = { , -2, -1, 0, 1, 2, } The rationals, which can be expressed as fractions of integers, are indicated by Q, while the real numbers, described through their decimal representation, are represented by R The symbol ∈ signifies 'belongs to'; for example, 1 ∈ N indicates that 1 is a natural number Additionally, the notation x ∈ R: x² = 2 refers to the set of real numbers whose square equals 2.
{x∈R: −1< x≤2} is the set of real numbers strictly larger than−1 and smaller than or equal to 2.
A shorter notation for such a set is(−1,2] The set x 2 :x∈ [−1,1] is the set of squares of numbers in[−1,1] The notation
{−1,2,4} designates a set with three elements:−1, 2, and 4 The empty set will be denoted by∅.
In mathematics, foundational principles known as axioms serve as the starting point for various branches such as geometry and probability These axioms are accepted without proof and form the basis for deriving all other mathematical truths For example, Euclidean geometry is built upon five fundamental axioms.
The Peano axioms provide a foundational framework for defining the set of natural numbers (N), the addition operation, and the order relation (0, by C1xx < xy Similarly, sincex < yandy >0, by C1 xy < yy By R2,x 2 < y 2
Letx =1/x If we hadx 0.
By R4xy >0, and so by C4 xy 1 >0 By C1,
1.1 The Algebra of the Reals 3
We now turn to the function absolute value We define
The following simple lemma is very useful.
Lemma Leta >0 be a real number We have|x| ≤aif and only if−a≤x≤a.
We prove the lemma It is an ‘If and only if’ statement, which means that we have two implications to prove.
For the direct implication, assume that|x| ≤a There are two cases Ifx ≥0,
|x| =x andx≤a Sincex≥0, we must have−a≤x Thus,−a≤x≤a On the other hand, ifx≤0,|x| = −x and−x≤a, that is,x≥ −a Sincex≤0, we have x≤a In both cases−a≤x≤a, and the direct implication is proved.
For the converse, assume now that−a≤x≤a If|x| =x, then|x| ≤a If|x| −x, since−a≤x, we havea≥ −x anda≥ |x| In both cases |x| ≤a, and the lemma is proved.
For any real numbersaandb, we have
For any realsaandb, we have
By adding the inequalities we get
According to the lemma above, this implies that
This completes the proof of the triangle inequality.
We now turn to an important algebraic identity We define thenth power of a real aby a 1 =a and a n + 1 =a n a for alln≥1.
For any natural numbernand any real numbersaandb, we have that a n −b n =(a−b) n − 1 k = 0 a n − 1 − k b k
Forn=2 andn=3, these are the well-known identities a 2 −b 2 =(a−b)
Forn≥4, the identity is a n −b n =(a−b) a n− 1 +a n− 2 b+ ã ã ã +ab n− 2 +b n− 1
We now prove it We have
The difference between the two sums is a n +a n − 1 b+ ã ã ã +a 2 b n − 2 +ab n − 1 − a n − 1 b+a n − 2 b 2 + ã ã ã +ab n − 1 +b n
All the terms cancel excepta n −b n This proves the identity above.
1 (a) Describe in words the set x∈Q:x 2 a We can setq=0,r=a, and we have a=qb+r with 0≤r < b.
If \( n_0 > 1 \), we define \( q = n_0 - 1 \) and \( r = a - bq \) Since \( n_0 \) is the minimum of set \( A \), \( n_0 - 1 \) cannot belong to \( A \) for two reasons: it is either not a natural number or \( b(n_0 - 1) \leq a \) Given that \( n_0 > 1 \), \( n_0 - 1 \) is indeed a natural number Therefore, we have \( bq = b(n_0 - 1) \leq a \), which implies \( r = a - bq \geq 0 \) Additionally, since \( n_0 \) is in \( A \), it follows that \( a - bn_0 < 0 \), leading to the conclusion that \( r = a - b(n_0 - 1) = a - bn_0 + b < b \).
We have proved the existence ofrandq.
Every natural number \( n \) can be expressed in one of two forms: \( n = 2m + 1 \), indicating that the number is odd, or \( n = 2m \), which means the number is even.
We do the long division ofnby 2 There are positive integersmandrsuch that n=2m+r where 0≤r m - \frac{1}{2} \), thus confirming our assertion.
Example 1.10 The fundamental property does not hold for the rationals Consider
That is,Ais the set of all rationals strictly larger than√
(2 is inA) and is bounded below by √
2 Assume, by contradiction, that A has a greatest lower boundmin the rationals If m a.
To demonstrate the Archimedean property through proof by contradiction, we begin by assuming it is false This leads us to the conclusion that there exists a real number \( a \) such that for every natural number \( n \), the inequality \( n \leq a \) holds true.
If the Archimedean property does not hold, then the set of natural numbers \( N \) is bounded above by a real number \( a \) According to the fundamental property of real numbers, the nonempty set \( N \) must have a least upper bound, denoted as \( m \) Consequently, \( m - 1 \) cannot serve as an upper bound for \( N \), indicating that there exists a natural number \( n \) such that \( n > m - 1 \) This leads to the conclusion that \( n + 1 > m \).
That is,n+1 is a natural is larger thanm, which is an upper bound of the naturals.
We have our contradiction The Archimedean property must hold.
Intuitively it is clear that 1 is the least upper bound ofB We will now prove this. First, observe that for every naturaln, we have
In this analysis, we establish that 1 serves as an upper bound for the set B To demonstrate that 1 is the least upper bound, we will select any value m that is less than 1 and show that m cannot qualify as an upper bound for B According to the Archimedean property, there exists a natural number n such that n exceeds 1.
Therefore, 1−m >1/nandm 0 be two reals Then there exists an integern≥0 such that n≤x < (n+1).
By the Archimedean property, there is a naturalpsuch that p > x/.
The set A is a nonempty subset of the natural numbers, containing the element p According to the well-ordering principle, A has a minimum element, denoted as m, which implies that m−1 cannot be in A This leads to two scenarios: either m equals 1 or m is greater than or equal to 2 If m equals 1, then it follows that x must be less than a certain value Consequently, for n equal to 0, the inequality n ≤ x < (n + 1) holds true.
On the other hand, ifm≥2, then m−1 is a natural, and since it is not inA
Using thatmis inA, we get
That is, n≤x < (n+1) forn=m−1, and the lemma is proved.
We now prove that the rationals are dense in the reals Assume that 0≤a < b.
We need to find a rational betweena andb By the Archimedean property, there is a naturalq such that q > 1 b−a.
Using the lemma with = q 1 , there exists a naturalnsuch that n≤a < (n+1).
Letr= n + q 1 ;ris a rational, andr > a From the inequality q > 1 b−a we get b >1 q +a≥ 1 q +n q =r.
Power Functions
We start by stating easy but important inequalities.
I1 If 0≤a < b, thena n < b n for any natural numbern.
I2 If 0< a 1, thena n > afor any natural numbern≥2.
We now prove these inequalities We have b n −a n =(b−a) n − 1 k = 0 b n − 1 − k a k
Note that all the terms in the sum are positive or 0 Hence, the sum is larger than its first termb n− 1 >0 That is, n − 1 k = 0 b n− 1 −k a k ≥b n− 1
To prove I2 and I3, note that a n −a=a a n − 1 −1
By I1, ifa 0 This proves I3.
For any natural numbernand any real numbera≥0, there is a unique positive solutionx of the equationx n =a We call the solution thenth root ofaand denote it bya 1/n
The strategy to establish the existence and uniqueness of a positive solution for the equation \( x^n = a \) mirrors the approach used for the specific case when \( n = 2 \) Initially, we employ an algebraic identity to demonstrate uniqueness, followed by a proof of existence utilizing the fundamental property of real numbers.
1 This proof is a little long and may be omitted.
Lemma 1.1 Assume thatx≥0,y≥0, andx n =y n for some natural numbern. Thenx=y.
Ifx n =y n =0, thenx=y=0 and we are done Ifx n =y n >0, sincex≥0 and y≥0, we havex >0 andy >0 (why?) We have x n −y n =(x−y) n − 1 k = 0 x n − 1 − k y k =0.
Sincex andy are strictly positive, the sum in the right-hand side has only strictly positive terms Thus,x−y=0, and we have proved Lemma1.1.
Lemma 1.1 indicates that there can be at most one solution for the equation \( x^n = a \) When \( a = 0 \), it is evident that the only solution is \( x = 0 \) For the remainder of the existence proof, we consider the case where \( a > 0 \) and proceed to define the relevant set.
Note that sincea >0 we havec= a + a 1 0 By I2c n < c Note also thatc < a(why?), and soc n < c < a That is,cis inA, andAis nonempty. Observe now that ift > a+1, then by I1t n > (a+1) n Moreover, by I3(a+
1) n > a+1> a, and sot n > a Thus, ift > a+1, thent is not inA Therefore, a+1 is an upper bound ofA SinceAis bounded above bya+1 and nonempty, it has a least upper bound that we denote bym.
We want to show thatm n =a and therefore thatmis the unique solution of the equationx n =a We do a proof by contradiction We will need the following:
Lemma 1.2 Assume that 0< x < yand thatn≥2 is a natural Then
To prove Lemma1.2, we start with y n −x n =(y−x) n − 1 k = 0 y n − 1 − k x k
Given the condition 0 < x < y, substituting x with y in each term of the sum results in a larger value For every integer k from 0 to n−1, the inequality x^k ≤ y^k holds, with strict inequality for k ≥ 1 Therefore, we have y^(n−1−k) x^k ≤ y^(n−1−k) y^k = y^(n−1) Consequently, it follows that y^n − x^n < (y−x)^(n−1) k = 0 y^(n−1) = (y−x)^n y^(n−1), utilizing the fact that there are n identical terms in the sum This establishes the proof for Lemma 1.2.
Assume now thatm n < a Let 0< δ m n Thus,(m+δ)is inAbut is larger than the upper boundm This is a contradiction We cannot havea > m n
We now prove that we cannot havea < m n either, and therefore we havea=m n Assume by contradiction thata < m n Let 0< h < m Apply Lemma1.2to get m n −(m−h) n < nhm n− 1
Note thath >0 sincem n > a Note also that h=m n −a nm n − 1 < m n nm n − 1 =m n ≤m.
Hence,h < m We have nhm n − 1 =nm n − 1 m n −a nm n− 1 =m n −a.
In conclusion, we establish that for any \( a \geq 0 \), the equation \( x^n = a \) has a unique positive solution This is demonstrated by showing that if \( t > m - h \), then \( t^n > (m - h)^n > a \), implying that \( t \) cannot belong to set \( A \) Consequently, there are no real numbers above \( m - h \) in \( A \), indicating that \( m - h \) serves as an upper bound for \( A \) Since \( m \) is assumed to be the least upper bound, we arrive at a contradiction, reinforcing the uniqueness of the positive solution.
We now state some important rules for powers.
For any real numbersa >0,b >0 and natural numberssandt, we have:
Recall our definition ofa n : a 1 =a and a n + 1 =a n a for alln≥1.
The rules P1, P2, and P3 can be proved by induction For P1, we fix a naturals, and we do an induction ont Fort=1, by definition we have a s + 1 =a s a.
The formula holds for \( t = 1 \) Assuming it holds for \( t \), we have \( a^{s + t + 1} = a^{s + t} \cdot a = a^s \cdot a^t \cdot a = a^s \cdot a^t + 1 \), where the second equality is based on the induction hypothesis The other equalities derive from our definition of power Therefore, we prove \( P1 \) by induction for a fixed \( s \) and any natural \( t \) Since \( s \) can be any natural number, this establishes \( P1 \) for all natural numbers \( s \) and \( t \).
For P2, we also fixsand do an induction ont Note that a s 1
=a s , and the formula holds fort=1 Assume that it holds fort Letc=a s By definition, a s t+ 1
By the induction hypothesis, a s t a s =a st a s
By P1 we have a st a s =a st + s =a s(t + 1) , and thus P2 holds for allsandt.
P3 may be proved by induction onsand is left as an exercise.
We now turn to rational powers We have defined 3 1/5 as being the unique posi- tive solution of the equationx 5 =3 But what does 3 2/5 mean? We define
To ensure the validity of the second equality, we must confirm that 3 2/5 is equivalent to 3 4/10 This illustrates that a proper definition of 3r for a rational r is independent of the specific fraction representing r The following sections will address these important considerations.
For any positive rational numberr=n/m(wherenandmare natural inte- gers) and any real numbera≥0, we define a r a n 1/m a 1/m n
Ifris a negative rational number anda >0, thena r is defined as 1/a − r
For the definition above to make sense, we need to check that a n 1/m a 1/m n and that ifr=n/m=p/q, then a n 1/m a p 1/q
Letmandnbe natural numbers,a a positive real number, and letx=(a 1/m ) n and y=(a n ) 1/m Observe thaty m =a n (by definition of themth root of a n ) On the other hand, x m a 1/m n m a 1/m m n
=a n , where the second equality comes from P2: for positive integersnandm,(b n ) m (b m ) n =b nm Thus,x m =y m , and by Lemma1.1,x=y.
We now check that ifr=n/m=p/q, then a n 1/m a p 1/q
Letx be the right-hand side, andy the left-hand side By the definition of themth root ofa n we have x m =a n and by P2 x m q
Since n/m andp/q represent the same rational number, we have nq=mp, so x mq =y mq , and by Lemma1.1,x=y.
For any real numbera >0 and any rational numberssandt, the rules P1, P2, and P3 hold Moreover, we have
We will demonstrate P1 for positive rational exponents, which will subsequently enable us to establish P4 in the same context This approach will facilitate the proof of P1 across all scenarios, followed by P4 in all cases The proofs for P2 and P3 are more straightforward and are provided as exercises.
Let \( x = a^s + t \) and \( y = a^s a^t \), with \( s = \frac{m}{n} \) and \( t = \frac{p}{q} \), where \( m, n, p, q \) are natural numbers We can express \( x \) as \( a^{mq + np} \cdot a^{mq} + np \cdot \frac{1}{nq} \) based on the definition of a rational power This leads to the equation \( x \cdot n^q = a^{mq + np} \) Additionally, using the property of natural powers, we can rewrite \( y \) as \( a^{s \cdot n^q} \cdot a^{t \cdot n^q} \) Consequently, we find that \( y \cdot n^q = a^{mq} \cdot a^{np} = x \cdot n^q \) The final equality confirms that \( x = y \), thus proving the property for positive rational powers.
We now prove P4 for positive rationalssandt Assume thats > t Then by P1, for positive rationals, we have a s − t a t =a s − t + t =a s
We now turn tos < t Then a t − s a s =a t and a t − s =a t a s
Recall that by definition a − r = 1 a r for all rationalsr Thus,
This proves P4 for positive rationalssandt.
We now prove P1 for any rationals Assume thats >0 andt >0 By P4, for positive rationals, we have a s − t =a s a t =a s a − t
This proves P1 when one rational exponent is positive and the other negative When they are both negative, we use P1 for positive rational exponents to get a − s − t = 1 a s + t = 1 a s a t = 1 a s
This demonstrates that P1 holds true for both negative exponents and any two rational exponents Utilizing this principle, we can establish P4 in all scenarios According to P1, for any rational numbers s and t, the equation \( a^{s - t} = a^{s - t + t} = a^s \) is validated.
This proves P4 for any rational exponents.
Rational power functions are defined as relations between two sets, where each element in set A is assigned exactly one element in set B A function f has an inverse, denoted as f − 1, if it allows for the reversal of this assignment Specifically, for every element y in set B, there exists a unique solution x in set A that satisfies the equation f(x) = y.
If that is the case, we can define the inverse functionf − 1 by settingf − 1 (y)=x wherexis the unique solution off (x)=y By the definition of the inverse function, we have, for everyyinB, f (x)=f f − 1 (y)
For any rational number \( r > 0 \), the function \( f(x) = x^r \) is defined for positive real numbers When \( r \) is expressed as \( p/q \), where \( p \) and \( q \) are positive integers, it follows that for any \( x \geq 0 \), \( x^r = (x^p)^{1/q} \), representing the unique positive \( q \)th root of \( x^p \) Conversely, if \( r < 0 \), then \( x^r = 1/x^{-r} \), which is also uniquely defined for all \( x \geq 0 \).
Letr=0 be a rational, and lety≥0 be a real Is there a unique solution to x r =y?
Yes, the equation can be expressed as Letr = p/q, where p and q are positive integers By setting x = y^(1/r), we derive that x = y^(q/p), leading to the relationship x^p = y^q and y = x^r This demonstrates that a unique solution exists for the given equation.
We now summarize our findings.
Fig 1.1 These are the graphs of y = √ x , y = x and y = x 2
Letr=0 be a rational number The functionx→x r is defined on the positive real numbers It has an inverse functionx→x 1/r
Figure1.1represents the graphs of y=√ x,y=x, and y=x 2 A number of interesting facts are contained in this picture In particular, ifr xfor x 1 Ifr >1, thenx r < xforx xforx >1.
Letr >0 be a rational number The functionx→x r is strictly increasing on the positive real numbers That is, for any real numbers 0≤a < b, a r < b r
This is in fact an easy consequence of the same property for natural powers I1.
We first prove the property in the particular caser=1/n Assume that 0< a < b, and letnbe a natural number Ifa 1/n ≥b 1/n , then we would have, by I1, a 1/n n
, and soa≥b This is a contradiction Therefore,a 1/n < b 1/n
In the general case where \( l = m/n \) with \( m \) and \( n \) as natural numbers, and assuming \( 0 < a < b \), it follows that \( a^{1/n} < b^{1/n} \) Consequently, raising both sides to the power of \( m \) yields \( (a^{1/n})^m < (b^{1/n})^m \) This demonstrates that \( a^r < b^r \) for \( r > 0 \), thereby proving that the function \( x \to x^r \) is increasing on the positive real numbers.
We now give two more useful inequalities.
I4 Let 0< r a m if and only ifn > m.
7 Assume thata >1 is real and that the rationals s andt are such that s < t.
Prove that a s < a t (Raise both quantities to the appropriate power and use Exercise 6.)
8 Show that if the rational number r is negative, then the function x→x r is decreasing on the positive reals That is, show that if 0≤a < b, thena r > b r
10 Assume that the rational numberr >1.
(b) Find two rationalsr 1 andr 2 such thatr 1 0, there is a natural numberN (that usually depends on ) such that ifn≥N, then
|a n −a|< a n converges toais denoted by lim n →∞ a n =a The numberais called the limit of the sequencea n
In the study of sequences, it is important to focus on the limit as n approaches infinity A sequence that fails to converge is classified as divergent We will explore various examples to illustrate this concept.
Example 2.1 Consider a constant sequencea n =cfor alln≥1, wherecis a con- stant Then, of course,a n converges toc: take any >0, takeN=1; then for every n≥N, we have|a n −c| =0< Therefore, by definition,a n converges toc.
Example 2.2 Leta n =1/n Intuitively, it is clear thata n converges to 0 We now prove this Let >0 By the Archimedean property there is a naturalN >1/ For n≥N, we have 1/n≤1/N < (why?) Thus,
1/n= |a n −0|< for alln≥N, and we have proved thata n converges to 0.
R.B Schinazi, From Calculus to Analysis,
DOI 10.1007/978-0-8176-8289-7_2, © Springer Science+Business Media, LLC 2012
Example 2.3 Assume that the sequencea n converges to Letb n =a n + 1 Let us show thatb n converges to the same
Since the sequence \( a_n \) converges to a limit \( L \), for any \( \epsilon > 0 \), there exists a natural number \( N \) such that for all \( n \geq N \), the condition \( |a_n - L| < \epsilon \) holds Furthermore, if \( n \geq N \), then \( n+1 \geq N \), which implies \( |a_{n+1} - L| < \epsilon \) as well Consequently, this means that the sequence \( b_n \) also converges to the same limit \( L \).
In order to prove that a sequence converges to 0, the following is sometimes useful.
A sequencea n converges to 0 if and only if the sequence|a n |converges to 0.
Assume first thata n converges to 0 For any >0, there isNsuch that ifn≥N, then
|a n | −0< for alln≥N This proves that|a n |converges to 0.
Assume now that |a n | converges to 0 For any >0, there isN such that if n≥N, then
|a n −0|< for alln≥N This proves thata n converges to 0 and completes the proof.
Example 2.4 Assume thata n converges to , and letc be a constant Then ca n converges toc.
Ifc=0, thenca n is the constant sequence 0, and it converges to 0=0ã So the property holds whenc=0 Assume now thatc=0 Let >0; sincea n converges to , there is a naturalNsuch that ifn≥N, we have
Multiplying across this inequality by|c|, we get
This proves thatca n converges toc, and we are done.
2.1 Sequences 35 The next example shows that not all oscillating sequences diverge.
Example 2.5 Leta n =(−1) n /n Show thata n converges to 0.
Observe that|a n | =1/nconverges to 0 Thus,a n converges to 0.
We now turn to the notion of bounded sequence
A sequencea n is said to be bounded if there exists a real numberKsuch that
There is an important relation between bounded sequences and convergent se- quences.
If a sequence converges, then it must be bounded.
We now prove that a convergent sequence is bounded Assume that the sequence a n converges to some limita Take =1; sincea n converges toa, there is a natural
N such that ifn≥N, then |a n −a|< =1 Thus, by the triangle inequality we have
The inequalities indicate that the sequence remains bounded for n ≥ N Now, we will address the case when n < N It is important to note that any finite set of real numbers will always have both a maximum and a minimum value.
, and letK=max(M,1+ |a|) Then, we claim that
For ifn < N, then|a n |< M≤K, while ifn≥N, then|a n |1 Consider the sequenceb n =c n forn≥0 Let us show thatb n does not converge.
Let|c| =1+awherea= |c| −1>0 By Bernoulli’s inequality we have
Note that the sequence 1+na is not bounded (why?), and therefore b n is not bounded either Therefore, it cannot converge.
A bounded sequence does not necessarily converge In order to show that a se- quence does not converge, the following notion of subsequence is quite useful.
In a strictly increasing sequence of natural numbers defined as 1 ≤ j1 < j2 < < jn, a new sequence of real numbers can be formed from an original sequence, denoted as an This new sequence, represented as ajn, is known as a subsequence of an, highlighting that all terms ajn are derived from the original sequence an.
Useful examples of strictly increasing sequences of natural numbers arej n =n, j n =2n,j n =2n+1, andj n =2 n
A sequencea n converges toaif and only if all the subsequences ofa n con- verge toa.
The proof is straightforward Assume that the sequence \(a_n\) converges to \(a\), and let \(j_n\) be a subsequence of \(a_n\) For any \(\epsilon > 0\), there exists an \(N\) such that for all \(n \geq N\), the inequality \(|a_n - a| < \epsilon\) holds It can be demonstrated that \(j_n \geq n\) for all \(n \geq 1\) Consequently, if \(n \geq N\), then \(j_n \geq n \geq N\), which implies \(|a_{j_n} - a| < \epsilon\) Therefore, the subsequence \(a_{j_n}\) also converges to \(a\).
This proves that any subsequence ofa n converges toa.
We now prove the converse We assume that all subsequences ofa n converge toa Buta n is a subsequence of itself (takej n =n), and thusa n converges toaand we are done.
We use the preceding result in the next example.
The sequence defined by \( a_n = (-1)^n \) does not converge, as it oscillates between -1 and 1 To demonstrate this, we can analyze its subsequences: the subsequence \( a_{2n} \) is constant and converges to 1, while the subsequence \( a_{2n + 1} \) is also constant but converges to -1 Since these two subsequences converge to different limits, this confirms that the sequence \( (-1)^n \) does not converge.
The following principle is sometimes quite useful.
Assume that the three sequencesa n ,b n , andc n are such that a n ≤b n ≤c n and that the sequencesa n andc n converge to the same limit Thenb n con- verges to as well.
The result above not only proves the convergence ofb n , but it also gives the limit.
We now prove it Recall that
We start by writing thata n andc n converge to For any >0, there are naturals
N 1 andN 2 such that ifn≥N 1 , then
DefineN=max(N 1 , N 2 ), so that the two double inequalities above hold forn≥N. Thus, forn≥N, using thatb n is squeezed betweena n andc n , we have
That is,b n converges to , and the squeezing principle is proved.
Next we give an application of the squeezing principle.
Example 2.8 Assume thata n converges to 0 and that b n is bounded Thena n b n converges to 0.
Sinceb n is bounded, there isBsuch that|b n |< Bfor alln Then,
That is, the sequence|a n b n |is squeezed between the constant sequence 0 and the sequenceB|a n | But they both converge to 0 (why?) By the squeezing principle,
|a n b n |converges to 0, and so doesa n b n
Example 2.9 Consider the sequencea n =n r whereris a rational number Discuss the convergence ofa n in function ofr.
Ifr=0, thena n is the constant sequence 1, and it converges to 1.
Assuming r > 0, we can prove by contradiction that the sequence \( a_n = n^r \) is unbounded If we assume it is bounded, there exists a constant \( B > 0 \) such that \( n^r < B \) for all natural numbers \( n \) Given that \( 1/r > 0 \), the function \( x \to x^{1/r} \) is increasing, which implies \( n < B^{1/r} \) for all natural numbers \( n \) This leads to a contradiction with the Archimedean property, demonstrating that the sequence \( a_n \) cannot be bounded Therefore, if \( r > 0 \), the sequence \( a_n \) does not converge.
Letr >0 We now show thatn − r converges to 0 Take any >0 and letN >
This proves thatn − r converges to 0.
The following is a useful property of least upper bounds.
Least upper bound and sequences
Let A be a subset in the reals with a least upper boundm There exists a sequencea n inAthat converges tom.
The important part of this property is that the sequence is inA The same property holds for the greatest lower bound, and the proof is left to the reader.
To demonstrate the property, we establish that since \( m \) is the least upper bound of set \( A \), the value \( m-1 \) cannot serve as an upper bound for \( A \) This implies the existence of at least one element \( a \) in \( A \) such that \( a > m-1 \), which we denote as \( a_1 \) Furthermore, since \( m-1/2 \) also fails to be an upper bound of \( A \), there exists another element \( a \) in \( A \) such that \( a > m-1/2 \).
We pick one sucha, and we call ita 2 More generally, for every naturaln,m−1/n is not an upper bound ofA, and we may picka n inAsuch that a n > m−1/n.
Hence, there exists a sequencea n inAsuch thata n > m−1/nfor alln≥1 On the other hand, since the sequencea n is inA, we must havea n ≤mfor alln≥1 Thus, m−1/n < a n ≤m.
The sequence \( a_n \) converges to \( m \), as demonstrated by the convergence of \( m - \frac{1}{n} \) to \( m \) This conclusion follows from the application of the squeezing principle and the properties of limits that will be discussed in the subsequent section.
1 (a) Leta n be a sequence of reals converging to Letb n =a n− 1 Show thatb n converges to as well.
(b) State a generalization of the result in (a) and prove your claim.
2 Let 1≤j 1 < j 2 0.)
5 Assume thata n converges toa Prove that for any >0,a n is in(a−, a+) for allnexcept possibly finitely many.
(a) Show that there isN such that ifn≥N, thena n 1/2.
To modify the proof of the squeezing principle for the sequences \( a_n \), \( b_n \), and \( c_n \), where \( a_n \leq b_n \leq c_n \) for \( n \geq 1000 \) and both \( a_n \) and \( c_n \) converge to the same limit, one can assert that since the bounds converge to the same limit, the sequence \( b_n \) must also converge to that limit Therefore, by applying the squeezing principle, it follows that \( \lim_{n \to \infty} b_n = \lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n \) This ensures that the conclusion of the squeezing principle remains valid under these conditions.
8 LetAbe a subset in the reals with a greatest lower bound k Show that there exists a sequencea n inAthat converges tok.
9 Assume thata n is positive and converges to a limit
(a) Assume that =0 Show that√ a n converges to 0.
(b) Assume that >0 Show that√ a n converges to√
10 Leta n be a sequence of reals, and be a real Letb n = |a n − | Show thata n converges to if and only ifb n converges to 0.
Assume that the sequence \( a_n \) takes values from the set \{0, 1, 2\} for every \( n \geq 1 \) and converges to 1 We need to demonstrate that there exists a natural number \( N \) such that for all \( n \geq N \), \( a_n \) equals 1.
12 Consider a sequencea n that takes values in the naturals.
(a) Give an example of such a sequence.
(b) Show thata n converges if and only if it is stationary, that is, if and only if there is a naturalN such that ifn≥N, thena n =a N
(a+b) n n k = 0 n k a k b n − k where all n k are natural numbers and n
14 (a) Assume thata n does not converge toa Show that there exists an >0 and a subsequencea j n such that
Let \( b_n \) be a sequence of real numbers Suppose there exists a limit \( b \) such that every subsequence of \( b_n \) contains a further subsequence that converges to \( b \) To prove by contradiction, assume that \( b_n \) does not converge to \( b \) This implies that there exists an \( \epsilon > 0 \) such that for infinitely many \( n \), \( |b_n - b| \geq \epsilon \) Consequently, we can construct a subsequence of \( b_n \) that does not converge to \( b \), contradicting the initial assumption Therefore, we conclude that \( b_n \) must converge to \( b \).
Monotone Sequences, Bolzano–Weierstrass Theorem, and
We start with the notion of monotone sequence.
A sequence \( a_n \) is classified as increasing if each term is greater than or equal to the preceding term, meaning \( a_{n+1} \geq a_n \) for all natural numbers \( n \) Conversely, a sequence is considered decreasing if each term is less than or equal to the previous term, represented as \( a_{n+1} \leq a_n \) for every natural number \( n \) Sequences that are either increasing or decreasing are referred to as monotone sequences.
The sequencea n =1/nis decreasing The sequenceb n =n 2 is increasing, and the sequence c n =(−1) n is not monotone The following notions are crucial for monotone sequences.
A sequence \( a_n \) is considered bounded below if there exists a real number \( m \) such that \( a_n > m \) for every \( n \) Similarly, a sequence \( a_n \) is deemed bounded above if there is a real number \( M \) such that \( a_n < M \) for all \( n \).
We are now ready to state an important convergence criterion for monotone se- quences.
An increasing sequence converges if and only if it is bounded above A de- creasing sequence is convergent if and only if it is bounded below.
We demonstrate that an increasing sequence \( a_n \) converges if it is bounded above Since a convergent sequence is necessarily bounded, we confirm one direction of the proof Conversely, if the increasing sequence \( a_n \) is bounded above by a number \( M \), we can analyze the set \( A \) formed from its elements to establish the convergence of the sequence.
The set A has an upper bound M and is not empty, which ensures the existence of a least upper bound for A due to the fundamental property of real numbers We will demonstrate that the sequence a_n converges to this least upper bound For any ε > 0, since M is the least upper bound, the value M - ε cannot serve as an upper bound, indicating that there exists at least one element in A that is strictly greater than M - ε.
Therefore, there isN such thata N > − Note that up to this point we have not used thata n is an increasing sequence We now do Taken≥N Then a n ≥a N > −.
2.2 Monotone Sequences, Bolzano–Weierstrass Theorem, and Operations 41
Since is larger than alla n , we have
In this analysis, we demonstrate that the sequence \( a_n = a^{1/n} \) converges, specifically for \( a > 1 \) We establish that \( a_{n+1} < a_n \), indicating that the sequence is decreasing Additionally, since \( a^{1/n} > 1 \) for all \( n \), and recognizing that the function \( x \to x^{1/n} \) is increasing, we conclude that \( a_n \) is not only decreasing but also bounded below by 1 Therefore, the sequence converges.
To determine the limit of the sequence \( a_n \), we analyze the subsequence \( a_{2n} \), which must also converge It is established that \( a_{2n} = a_{2n-1} = (a_n)^{1/2} \) converges to \( \frac{1}{2} \) (refer to Exercise 9 in Section 2.1) Consequently, we find that \( L = \frac{1}{2} \) The limit \( L \) can either be \( 0 \) or \( 1 \), but it cannot be both.
0 (why not?), therefore it is 1.
In the given example, let \( c \) be a constant in the interval (0,1) and define the sequence \( a_n = c^n \) It is demonstrated that \( c^n \) converges to 0 Since \( c < 1 \), it follows that \( a_{n+1} < a_n \), indicating that the sequence \( a_n \) is strictly decreasing Additionally, \( a_n \) is positive, thus it is bounded below by 0 Consequently, the sequence converges to some limit \( L \) As \( a_{n+1} \) also converges to the same limit, and given that \( a_{n+1} = c a_n \), we can conclude that \( c a_n \) converges to \( c \) Therefore, both \( a_{n+1} \) and \( c \) converge to the same limit \( L \), leading to the conclusion that \( L = c \).
Therefore, eitherc=1 (but we know thatc n k such that a n k+1 > a n k
That is, we may construct inductively a strictly increasing sequence whenAis finite. Note that in caseAis empty, the same construction works forN=1.
Assume now thatAis infinite SinceA is a nonempty set of naturals, it has a minimumn 1 Letn 2 > n 1be another element inA Sincen 1is inA, we have a n 2 ≤a n 1
Assume that we have foundn 1 < n 2 n k Using thatn k is inA, we have a n k + 1 ≤a n k
That is, we may construct inductively a decreasing sequence whenAis infinite This completes the proof of the lemma.
The Bolzano-Weierstrass theorem can be easily proven by considering a bounded sequence \( a_n \) According to the lemma, this sequence contains a monotone subsequence \( a_{n_k} \) Since \( a_n \) is bounded, the subsequence \( a_{n_k} \) is also bounded and monotone Consequently, it converges, thereby completing the proof of the theorem.
Example 2.12 Show that ifa n is a sequence in[0,1], then it has a subsequence that converges.
Since for alln≥1,a n is in[0,1], we have|a n | ≤1 Hencea n is bounded, and by Bolzano–Weierstrass, it has a convergent subsequence.
We now turn to the operations on limits.
Assume that the sequencesa n andb n converge, respectively, toaandb Then
(iii) Assume thatb n is never 0 and thatbis not 0 Then n lim→∞ a n /b n =a/b.
(iv) Assume that for alln,a n ≤b n Thena≤b.
2.2 Monotone Sequences, Bolzano–Weierstrass Theorem, and Operations 43
We prove the statements above Sincea n andb n converge, for any >0, there are natural numbersN 1 andN 2 such that ifn≥N 1 , we have
TakeN=max(N 1 , N 2 )so that both statements above hold forn≥N and use the triangle inequality to get
In this section, we address the case where b equals zero, utilizing the findings from Example 2.8 in Section 2.1 Given that the sequence a_n converges, it is also bounded We have previously established that the product of a bounded sequence and a sequence that converges to zero will also converge to zero.
We now assume thatb=0 We have
Sincea n converges, it is bounded, so there isA >0 such that|a n |< Afor everyn.
Take >0; there isN 1 such that ifn≥N 1 , then
There is alsoN 2such that ifn≥N 2, then
We use the two preceding inequalities in (2.1) to get that ifn≥max(N 1 , N 2 ),
This completes the proof of (ii).
To prove (iii), it is enough to prove that 1/b n converges to 1/b We can then use (ii) to show thata n /b n =a n (1/b n )converges toa(1/b)=a/b We start with
The only difficulty is to show thatb n is bounded away from 0 Sinceb n converges tob, there is a natural numberN 1such that ifn≥N 1, then
|b n −b| ≤ |b|/2, where we are using the definition of convergence with = |b|/2>0 (sinceb=0).
By the triangle inequality we get, forn≥N 1,
On the other hand, sinceb n converges tob, for any >0, there isN 2such that if n≥N 2, then
Therefore, forn≥max(N 1 , N 2 ), we have
That is, 1/b n converges to 1/b, and the proof of (iii) is complete.
For (iv), we do a proof by contradiction Assume thata > band let = a−b 2 >0. Sincea n converges toa, there existsN 1 such that ifn≥N 1 , then
In particular,a n > a− = a + 2 b ifn≥N 1 On the other hand, sinceb n converges tob, there existsN 2such that ifn≥N 2, then
In particular,b n < b+ = a + 2 b ifn≥N 2 Thus, ifn >max(N 1 , N 2 ), then b n m, thena n ≥a m
3 Prove that a decreasing sequence converges if and only if it is bounded below.
4 Assume that a n is an increasing sequence,b n is a decreasing sequence, and a n ≤b n for alln≥1.
Series
(b) If in addition to the previous hypotheses, we have thatb n −a n converges to 0, prove thata n andb n converge to the same limit.
5 Assume thata n < b n for all naturalsn Assume thata n andb n converge toa andb, respectively Is it true thata < b? Prove it or give a counterexample.
6 Assume thata n converges to and is bounded below bym Show that ≥m.
7 Show that for any a >0, a 1/n converges to 1 (you may use the fact that the result has been proved fora >1 in Example2.10).
8 Suppose thata n is increasing Show thata n converges if and only if it has a subsequence which converges.
(a) Show that there isN such forn≥N, we havea n in(0,1).
(b) Is the statement in (a) true if we replace(0,1)by[0,1)?
10 We define a sequencea n recursively by settinga 1=2 and a n + 1 =2a n −1 a n
(a) Compute the first five terms of this sequence.
(b) Show that for alln≥1, ifa n >1, thena n + 1 >1.
(c) Use (b) to show that the sequencea n is well defined and thata n is bounded below by 1.
(e) Show thata n converges to some limit ≥1.
11 Assume thata n −aconverges to 0 Show thata n converges toa.
12 Assume that|a n −a| ≤1/nfor alln≥1 Show thata n converges toa.
13 Assume thata n −b n converges and thatb n converges Show thata n converges.
14 Consider a sequencea n that converges toa >0 Show that there is a naturalN such that ifn≥N, thena n >0.
15 Assume that|c|0 such that(a−, a+)⊂O.
A series, as we are going to see, is the sum of a sequence.
Leta n be a sequence of real numbers Define the sequenceS n by
S n is the sequence of partial sums The series ∞ k = 1 a k is said to converge if the sequenceS n converges If that is the case, then the limit ofS n is denoted by ∞ k = 1 a k
The series above starts atk=1, but it may actually start at any positive or 0 integer A series that does not converge is said to diverge.
Example 2.14 Letrbe a real number in(−1,1), anda n =r n Recall that for any real numbersa,band natural numbern, we have a n + 1 −b n + 1 =(a−b) a n +a n − 1 b+ ã ã ã +ab n − 1 +b n
By Example2.3in Sect.2.1we know thatr n + 1 converges to 0 asngoes to infinity for|r| N Then
The convergence of the series ∞ n = 0 b n indicates that the sequence T n is bounded, specifically by some upper limit K Notably, since N is a constant, the difference S N − T N remains fixed and does not vary with n Consequently, the sequence S n is constrained above by the expression S N − T N + K.
On the other hand,S n + 1−S n =a n + 1≥0, and thusS n is an increasing sequence.
An increasing sequence which is bounded above must converge, and (1) is proved. Now we turn to (2) Assume that S n diverges We use the inequality proved above:
As already observed,S n is increasing; since it diverges, it cannot be bounded above. Thus,T n is larger thanS n −S N +T N , which is not bounded above (why?), and so
T n cannot be bounded either Therefore,T n diverges, and (2) is proved.
Example 2.17 Leta n ≥0 and assume that the series ∞ n = 1 a n diverges We show that ∞ n = 1
We consider two cases Ifa n does not converge to 0, then√ a n does not converge to 0 either (why?), and by the divergence test the series ∞ n = 1
Ifa n converges to 0, there isNsuch that ifn≥N, then
In particular, 0≤a n 0 such that a b n n converges to a strictly positive number Then, the series ∞ n = 0 a n and
∞ n = 0 b n both converge or both diverge.
We now prove the limit comparison test Assume thata n /b n converges to some
>0 By the definition of convergence with =/2, there isNsuch that ifn≥N, a n b n −
Assume that the series ∞ n = 0 a n converges Since b n 2
3a n forn≥N, the series ∞ n = 0 b n diverges by the comparison test.
To finalize the proof, we need to demonstrate that the convergence of the series ∞ n = 0 b n implies the convergence of the series ∞ n = 0 a n, and similarly, that the divergence of ∞ n = 0 b n leads to the divergence of ∞ n = 0 a n Notably, the ratio b n / a n converges to a strictly positive real number, specifically 1, indicating that a n and b n have symmetric roles in this context This establishes the validity of the limit comparison test.
Example 2.18 Does the series ∞ n = 2 n n 2 − 1converge?
Leta n = 1 n andb n = n 2 n − 1 It is easy to check thata n /b n converges to 1 Since the series ∞ n = 2 a n diverges, so does the series ∞ n = 2 b n by the limit comparison test.
Example 2.19 Leta≤1 be a real number The series
Leta n = n 1 andb n = n 1 a Both sequences are positive, and sincea≤1, a n ≤b n for alln≥1.
We have already shown that the series ∞ n = 1 a n diverges, so by the comparison test (withN=1) the series ∞ n= 1 b n diverges as well.
We still need to analyze the casea >1 for the series ∞ k = 1 1 k a In this case, 1 k> 1 k a
1 k diverges, this comparison is useless We need another test.
Consider a decreasing and positive sequencea n The series
∞ k = 1 a k converges if and only if the series
2 k a 2 k be the two partial sums.
Assume first that the series ∞ k = 1 a k converges and therefore that the sequence
Since the sequencea n is decreasing for every naturalj in[2 k − 1 +1,2 k ], we have a j ≥a 2 k Thus,
2 k j = 2 k − 1 + 1 a 2 k =2 k − 1 a 2 k , where we use that in[2 k − 1 +1,2 k ], there are 2 k − 1 naturals (why?) We sum the preceding inequality to get n k = 1
SinceS n converges, it must be bounded by someK Thus,
SoT n is bounded above Since it is also increasing (why?), it must converge We have proved the direct implication.
Assume now thatT n converges Consider again
Since the sequencea n is decreasing for every naturalj in[2 k− 1 +1,2 k ], we have a j ≤a 2 k − 1 Thus,
2.3 Series 53 and sinceS 1 =a 1 , we have
Butn 0 be a real number and consider the series ∞ k = 1
The series diverges for \( p \leq 1 \), but the Cauchy test can help us determine convergence in all cases To apply this test, we first verify its hypotheses The sequence \( a_k = \frac{1}{k^p} \) is positive and decreasing for all \( p > 0 \).
(2 k ) p 2 − p k and that the series of general term
2 k a 2 k 2 − p + 1 k is a geometric series with ratior=2 − p + 1 In particular,r 1.
By the Cauchy test, the series ∞ k= 1 1 k p converges if and only ifp >1.
We now have a complete result for series of the type ∞ k = 1 1 k p for realp Since this is an important class of series, we summarize our results below. p Test
1 k p converges whenp >1 and diverges whenp≤1.
(a) ComputeS n = n k = 1 a k in function ofn(observe thata n =1/n−1/(n+ 1)).
(b) Show that the series ∞ k = 1 a k converges by computing its sum.
4 Leta n ≥0 andb n =√ a n Show thata n converges to 0 if and onlyb n converges to 0.
5 (a) Leta n ≥0 be such that ∞ k = 1 a k converges Show that ∞ k = 1 a 2 k converges as well.
(b) Is the converse of (a) true?
(c) In (a) you have shown that if ∞ k = 1 a k converges, then ∞ k = 1 f (a k )con- verges forf (x)=x 2 Find other functionsf for which this is true.
6 Leta n >0 Show that ∞ k = 1 a k converges if and only if
7 Show that ∞ k = 1 a k converges if and only if ∞ k = 100 a k converges.
8 Assume that the sequencesa n ≥0 and b n >0 are such that a n /b n converges to 0.
(a) Show that if the series ∞ k = 1 b k converges, so does ∞ k = 1 a k (b) Show that if the series ∞ k = 1 a k diverges, so does ∞ k = 1 b k (c) Show by exhibiting counterexamples that the converses of (a) and (b) do not hold.
9 The sequencea n is said to go to+∞if for any realA >0, there is a natural numberN such that ifn≥N, thena n > A.
(a) Give an example of a sequence that goes to+∞ Prove your claim. (b) Show that a sequence that goes to+∞cannot be bounded.
(c) Is it true that any unbounded positive sequence goes to infinity?
10 Assume that the sequencesa n ≥0 andb n >0 are such thata n /b n goes to infin- ity (see the definition in Exercise 9).
(a) Show that if the series ∞ k = 1 a k converges, so does ∞ k = 1 b k (b) Show that if the series ∞ k = 1 b k diverges, so does ∞ k = 1 a k (c) Show by exhibiting counterexamples that the converses of (a) and (b) do not hold.
11 Assume that the sequencesa n ≥0 andb n ≥0 are such that ∞ k = 1 a k converges andb n is bounded Prove that ∞ k= 1 a k b k converges.
12 Assume that a n ≥0 and b n ≥0 for all n≥1 Suppose that ∞ k = 1 a k and
13 Prove that in[2 k− 1 +1,2 k ]there are 2 k− 1 naturals.
14 (a) Show that ifa n converges, thena n −a n − 1 converges to 0.
(b) Is the converse of (a) true? Prove it or give a counterexample.