For any given  &gt

Example 0.1. Prove that lim

n→∞

1 n = 0.

Proof. For any given  > 0, we choose

N = 1



 + 1.

We know N_ > 1/. Then for any n ≥ N_, we have 

   1 n− 0

= 1 n ≤ 1

N

< 1 1/ = .

This prove our assertion by definition. 

Example 0.2. Prove that lim

n→∞

1 2ⁿ = 0.

Proof. Let  > 0. If  > 1/2, we take N_= 1. Then 1

2ⁿ < , ∀n ≥ N_ = 1.

If 0 <  ≤ 1/2, we take N = [log₂1/] + 1. Then for all n ≥ N, 2ⁿ≥ 2^N > 2^log2(1/)= 1

. (Here we use a^logab= b.)

1 2ⁿ ≤ 1

2^N < 1 1/ = .

 Example 0.3. Prove that lim

n→∞

n²− n + 5 3n²+ n + 7 = 1

3. Proof. Observe that for any n ≥ 1,

n²− n + 5 3n²+ n + 7 −1

3 = −4n + 8 3(3n²+ n²+ 7).

For n ≥ 1, 3n² + n + 7 > 3n². Using triangle inequality, we also have | − 4n + 8| ≤

| − 4n| + |8| = 4n + 8 for n ≥ 1. Therefore for any n ≥ 1, combining all the inequalities we have so far,

n²− n + 5 3n²+ n + 7− 1

3    

≤ 4n + 8 9n² . If we require that n ≥ 8, we have a further estimate

n²− n + 5 3n²+ n + 7−1

3    

≤ 5n 9n² = 5

9n. Now let us prove the statement.

1

For any  > 0, we take

N = [5 9] + 9.

Then for any n ≥ N_, n ≥ 8 and n > 5/. Hence for n ≥ N_, we have 

n²− n + 5 3n²+ n + 7 −1

3    

≤ 5 9n ≤ 5

9n < .

This proves our assertion.

 Proposition 0.1. A convergent sequence is a Cauchy sequence.

Proof. Suppose (an) is convergent to a. For any  > 0, there exists N ∈ N such that

|a_n− a| < /2 whenever n ≥ N_. Using triangle inequality, we find that for any n, m ≥ N_,

|a_n− a_m| ≤ |a_n− a| + |a_m− a| <  2 +

2 = .

 Theorem 0.1. (Completeness of Rⁿ) A sequence of real numbers is convergent if and only if it is a Cauchy sequence.

Example 0.4. Prove that the sequence (sn) defined below is a Cauchy sequence and hence it is convergent.

sn= 1 + 1

2² + · · · + 1

n², n ≥ 1.

Proof. Suppose n > m ≥ 1. Then s_n− s_m= 1

(m + 1)² + 1

(m + 2)² + · · · + 1 n². For k ≥ 2, k²> k(k − 1). For k ≥ 2,

1

k² < 1

k(k − 1) = k − (k − 1) k(k − 1) = 1

k − 1 − 1 k. Using this observation, we know

0 < s_n− s_m = 1

(m + 1)² + 1

(m + 2)² + · · · + 1 n²

≤ 1

m(m + 1)+ · · · + 1 (n − 1)n

= 1

m − 1

m + 1



+ · · · +

 1

n − 1 − 1 n



= 1 m − 1

n

= 1 m.

In other words, we prove that for any n > m ≥ 1, we have 0 < s_n− s_m < 1

m.

For any  > 0, we take N_= [1/] + 1. For any n > m ≥ N_, we have

|s_n− s_m| = s_n− s_m< 1 m ≤ 1

N_ < 1 1/ = .

 Example 0.5. Show that for each n ≥ 1,

0 < 2^1/n− 1 < 1 n. This implies that lim

n→∞2^1/n= 1.

Proof. We use the formula

xⁿ− 1 = (x − 1)(xⁿ⁻¹+ · · · + x + 1).

Substitue x = 2^1/n into the above formula, we obtain

2^1/n− 1 = 1

2ⁿ⁻¹ⁿ + · · · + 2ⁿ¹ + 1 . Using 2^j > 1 for all j ≥ 0, we see

2ⁿ⁻¹ⁿ + · · · + 2ⁿ¹ + 1 > n.

Thus

1

2ⁿ⁻¹ⁿ + · · · + 2ⁿ¹ + 1 < 1 n.

 Example 0.6. Define

x_n= 1 + 1

1! + · · · + 1

n!, n ≥ 1.

Show that the sequence (xn) is increasing and bounded above. Hence (xn) is convergent.

Proof. For n ≥ 1,

x_n+1 = x_n+ 1

(n + 1)! > x_n. This shows that (xn) is increasing.

For k ≥ 2, k! ≥ k(k − 1). For n ≥ 2, xn= 1 + 1

1!+ 1 2!+ 1

3!+ · · · + 1 n!

< 1 + 1 + 1

1 · 2 + 1

2 · 3 + · · · + 1 (n − 1) · n

= 2 +

 1 −1

2

 + 1

2 −1 3

 + · · ·

 1 n − 1 − 1

n



= 3 − 1 n

< 3.

This shows that (x_n) is bounded above. By the monotone sequence property, (x_n) is

convergent. 

Lemma 0.1. A convergent sequence is always bounded.

Proof. Suppose (a_n) is convergent to a. Choose  = 1. There exists N = N_ ∈ N so that

|a_n− a| < 1 whenever n ≥ N. This implies

|a_n| ≤ 1 + |a|

whenever n ≥ N. Let

M = max{|a1|, |a₂|, · · · , |a_{N −1}|, 1 + |a|}.

Then for all n ≥ 1, |a_n| ≤ M. This implies that (a_n) is bounded.

 Theorem 0.2. Let (a_n) and (b_n) be sequences so that lim

n→∞a_n= a and lim

n→∞b_n= b. Then (1) lim

n→∞(an+ bn) = a + b.

(2) lim

n→∞ka_n= ka for k ∈ R.

(3) lim

n→∞(a_nb_n) = ab.

(4) lim

n→∞

a_n bn

= a

b if b 6= 0.

Proof. (1) Since (a_n) and (b_n) are convergent to a and b respectivekly, for any  > 0, we can choose N ∈ N so that

|a_n− a| < 

2, |b_n− b| <  2, whenever n ≥ N_. Using triangle inequality, for n ≥ N_,

|a_n+ b_n− (a + b)| ≤ |a_n− a| + |b_n− b| <  2 + 

2 = .

By definition, (an+ bn) is convergent to a + b.

(2) If k = 0, the proof is obvious. Assume k 6= 0. Since (a_n) is convergent to a, for any

 > 0, we can choose N∈ N so that

|a_n− a| < 

|k|

whenever n ≥ N_. Multiplying the above inequality by |k|, we see

|ka_n− ka| <  whenever n ≥ N.

(3) Observe that

a_nb_n− ab = a_nb_n− ab_n+ ab_n− ab = (a_n− a)b_n+ a(b_n− b).

Assume that a 6= 0. Since (bn) is convergent, it is bounded. There exists M > 0 so that

|b_n| ≤ M for all n ≥ 1. Since (a_n) is convergent to a and (b_n) is convergent to b, for any

 > 0, we can choose N∈ N so that

|a_n− a| < 

2M, |b_n− b| < 

2|a|, n ≥ N_. Hence for n ≥ N_,

|a_nb_n− ab| = |a_n− a||b_n| + |a||b_n− b|

≤ M |a_n− a| + |a||b_n− b|

< M · 

2M + |a| ·  2|a|

=  2 + 

2

= .

This shows that lim

n→∞anbn= ab.

When a = 0, we only need to show lim

n→∞a_nb_n = 0. Since lim

n→∞a_n = 0, for any  > 0, there exists N_ ∈ N so that for any n ≥ N, |a_n| < /M. Here M is as above, Thus

|a_nbn| = |a_n||b_n| < 

M · M = .

whenever n ≥ N. This proves our assertion. 

Corollary 0.1. Suppose (an) is convergent to a. For k ≥ 1,

n→∞lim a^k_n= a^k.

Proof. This can be proved by induction. The statement is true for k = 1 by assumption.

Suppose the statement is true for k = m, i.e. lim

n→∞a^m_n = a^m. Since (an) is convergent to a and (a^m_n) is convergent to a^m, by property (3), we find

n→∞lim a^m+1_n = lim

n→∞a_n· a^m_n = lim

n→∞a_n· lim

n→∞a^m_n = a · a^m= a^m+1.

Hence the statement is true for k = m + 1. By mathematical induction, the statement is

true for k ≥ 1. 

Lemma 0.2. Suppose (a_n) and (b_n) are convergent to a and b respectively and there exists N > 0 so that an≤ b_n for all n ≥ N. Then a ≤ b.

Proof. Since both (an) and (bn) are convergent, for  > 0, we can choose N> 0 such that for n ≥ N_,

a −  < an< a + , b −  < bn< b + .

Let n ≥ N and n ≥ N. Then

a −  < an< bn< b + .

We see that a < b + . Since  > 0 is arbitrary, we obtain a ≤ b.

 Proposition 0.2. Suppose (an) is convergent to a. Assume an≥ 0 for all n. For k ≥ 1,

n→∞lim

√k

a_n= √^k a.

Proof. Using above lemma, we know that a ≥ 0 since a_n≥ 0 for all n ≥ 1.

Assume a = 0. Since (an) is convergent to a = 0, we can choose N > 0 so that |an| < ^k for n ≥ N_. Thus

|√^k

an| < , whenever n ≥ N. This shows that lim

n→∞

√k

an= 0 = √^k a.

Assume that a > 0. Using the identity

x^k− y^k = (x − y)(x^k−1+ x^k−2y + · · · + xy^k−2+ y^k−1), we obtain

√k

a_n−√^k

a = a_n− a a

k−1

nk + · · · + a^k−1k . Since a_n≥ 0,

 ^k

√a_n−√^k a

= |a_n− a|

a

k−1

nk + · · · + a^k−1k

< |a_n− a|

a^k−1k . Since (a_n) is convergent to a, for any  > 0, we can choose N_> 0 so that

|a_n− a| < a^k−1k , whenever n ≥ N. Hence for n ≥ N_,

 ^k

√an−√^k a

< |a_n− a|

a^k−1k

< a^k−1k  a^k−1k

= .

This proves that lim

n→∞

√k

a_n= √^k a.

 Let (xn) be the sequence of numbers defined by

xn= 1 + 1 1! + 1

2!+ · · · + 1

n!, n ≥ 1.

We have already seen that (x_n) is convergent in the previous section. Let us denote

n→∞lim xn= x. Now, let (yn) be the sequence of numbers defined by

yn=

 1 + 1

n

n

, n ≥ 1.

Proposition 0.3. (y_n) is convergent.

Before proving Proposition 0.3, let us review the following two important facts.

Theorem 0.3. (A.G. inequality) Let a1, · · · , an be any nonnegative real numbers. Then a1+ · · · + an

n ≥ √ⁿ

a1· · · a_n.

Theorem 0.4. (Binomial Theorem) Let x, y be real numbers. Then

(x + y)ⁿ=

n

X

k=0

n k



x^ky^n−k.

Here n k



= n!

k!(n − k)!.

Let us choose a₁ = 1 and a_k= 1 + 1

n for 2 ≤ k ≤ n + 1. Then a₁+ · · · + a_n+1

n + 1 = 1 + n · 1 + ¹_n

n + 1 = n + 2

n + 1 = 1 + 1 n + 1. and

a1a2· · · a_n+1 =

 1 + 1

n

n

Using A.G. inequality, we see that

1 + 1

n + 1 > ⁿ⁺¹ s

 1 + 1

n

n

.

Taking n + 1-th power of both side of the inequality, we get

y_n+1=



1 + 1 n + 1

n+1

>

 1 + 1

n

n

= y_n.

Hence (y_n) is an increasing sequence. Using the binomial theorem, we see that y_n=

 1 + 1

n

n

= 1 +n 1

 1 n+n

2

 1

n² + · · · +n n

 1 nⁿ

= 1 + n · 1

n+n(n − 1) 2!

1 n² + · · ·

= 1 + 1 + 1 2!

 1 − 1

n

 + 1

3!

 1 − 1

n

  1 −2

n



+ · · · + 1 n!

 1 − 1

n



· · ·



1 −n − 1 n

 . Here we use the following simple equality:

n k

 1

n^k = n(n − 1) · · · (n − k + 1)

k! · 1

n^k = 1 k!

 1 − 1

n



· · ·



1 −k − 1 n

 . Notice that 1 − j

n < 1 for all 1 ≤ j ≤ n. Hence

n k

 1 n^k = 1

k!

 1 −1

n



· · ·



1 −k − 1 n



< 1 k!. Therefore

yn= 1 +n 1

 1 n +n

2

 1

n² + · · · +n n

 1

nⁿ < 1 + 1 1!+ 1

2!+ · · · + 1 n! = xn. We have already seen that xn< 3 for all n ≥ 1, and hence

yn≤ x_n< 3

for all n ≥ 1. This shows that (yn) is bounded. We conclude that (yn) is convergent by Theorem ??. We let us denote y = lim

n→∞y_n.

Theorem 0.5. Let (xn) and (yn) be as above and x and y be their limits respectively.

Then x = y.

For each 1 ≤ k ≤ n, yn= 1 + 1 +n

2

 1

n² + · · · +n k

 1 n^k +

 n k + 1

 1

n^k+1 + · · · +n n

 1 nⁿ

> 1 + 1 +n 2

 1

n² + · · · +n k

 1 n^k. Here we use the fact that

 n k + 1

 1

n^k+1 + · · · +n n

 1 nⁿ > 0.

Since we know that

n j

 1 n^j = 1

j!

 1 − 1

n



· · ·



1 −j − 1 n

 ,

we see that

n→∞lim

n j

 1 n^j = 1

j!. Hence taking n → ∞ of the following inequality

yn> 1 + 1 +n 2

 1

n² + · · · +n k

 1 n^k, we see that by Lemma ??,

y ≥ 1 + 1 + 1

2!+ · · · + 1 k! = xk

for all k ≥ 1. We see that

x = lim

k→∞x_k≤ y.

We obtain x ≤ y. Hence x ≤ y and y ≤ x. This implies x = y.

Definition 0.1. Let (xn) be the sequence defined as above. We denote e = lim

n→∞x_n. The number e is called the natural exponent.

We also obtain from the previous discussion that Theorem 0.6.

n→∞lim

 1 + 1

n

n

= e.

Remark. It follows from the definition that e < 3.

Theorem 0.7. (Sandwich Principle) Let (an) and (bn) and (cn) be sequences. Suppose there exists N ∈ N so that

an≤ b_n≤ c_n and limn→∞an= limn→∞cn= a. Then limn→∞bn= a.

Proposition 0.4. If (a_n) is a Cauchy sequence, then it is bounded.

Proof. Choose  = 1. Then we can find N so that for any n ≥ N , |an− a_N| < 1. Hence

|a_n| ≤ 1 + |a_N| for any n ≥ N . Let M = max{|a₁|, · · · , |a_{N −1}|, 1 + |a_N|}. Then |a_n| ≤ M for any n ∈ N.

 Let (a_n) be a Cauchy sequence. Then (a_n) is bounded. By Bolzano-Weierstrass theorem, (a_n) has a convergent subsequence (a_nk). Denote a = lim_k→∞a_nk.

Proposition 0.5. If (an) is a Cauchy sequence and it has a convergent subsequence, then it is convergent. More precisely, if (a_nk) is a subsequence of (a_n) convergent to a. Then (an) is also convergent to a.

Proof. Let (a_nk) be a subsequence which is convergent. Let a be its limit. Then given

 > 0, there exists K > 0 so that for any k ≥ N, |ank− a| < /2. Since (a_n) is a Cauchy sequence, for the same , there exists K_⁰ > 0 so that for any m, n ≥ K_⁰, |a_n− a_m| < /2.

Let us take N_ = max{K_, K_⁰}. Then for n, k ≥ N_, n_k≥ N_ and thus

|a_n− a| ≤ |a_n− a_nk| + |a_nk− a|

<  2+ 

2 = .

Hence (a_n) is convergent to a.

 Corollary 0.2. A sequence is convergent if and only if it is a Cauchy sequence.

Proof. We have already shown that every convergent sequence is a Cauchy sequence. Now, we need to show that every Cauchy sequence is convergent.

Suppose that (a_n) is a Cauchy sequence. Then it is bounded. Since every bounded sequence has a convergent subsequence, (a_n) has a convergent subsequence said (a_nk).

Suppose the limit of (an_k) is a. Then by the previous theorem, we find a is also the limit of (a_n). Hence (a_n) is convergent.

 Theorem 0.8. (Bolzano-Weierstrass theorem) Any bounded sequence of real numbers has a convergent subsequence.

Proof. Suppose (a_n) is bounded. Assume that a ≤ a_n≤ b for all n ≥ 1. Divide the interval I = [a, b] into two subintervals with equal length. At least one of the subintervals contains infinite many terms of the sequence (a_n). Choose one and denote it by I₁ = [α₁, β₁]. Choose one term an1 of the sequence (an) in I1. Now, α1 ≤ a_n1 ≤ β₁ and l(I1) = β1−α₁ = (b−a)/2.

Then we divide I₁ into two subinterval with equal length. Choose one so that it contains infinite many terms of the sequence and denote it by I2= [α2, β2]. Choose an2 ∈ I₂ so that n₂ > n₁. Inductively, we can choose a sequence of intervals I_k = [α_k, β_k] and a_nk ∈ I_k so that n_k+1 > n_k for all k and l(I_k) = β_k− α_k = (b − a)/2^k. Then (α_k) is a nondecreasing sequence bounded above by b while (βk) is a nonincreasing sequence bounded below by a.

Using the monotone sequence properties, both (α_k) and (β_k) are convergent. Denote the limits of (α_k) and (β_k) by α and β respectively. Since l(I_k) = (b − a)/2^k = β_k− α_k, and lim_k→∞α_k= α, and lim_k→∞β_k= β we find

0 = lim

k→∞

b − a

2^k = lim

k→∞(β_k− α_k) = lim

k→∞β_k− lim

k→∞α_k= β − α.

This shows that β = α. Denote them by a. Since ank ∈ I_k for all k, αk ≤ a_nk ≤ β_k for all k ≥ 1. By the Sandwich principle, lim_k→∞a_nk = a. This completes the proof of our

assertion. 

1. Supremum and Infimum

Remark: In this sections, all the subsets of R are assumed to be nonempty.

Let E be a subset of R. We say that E is bounded above if there exists a real number U such that x ≤ U for all x ∈ E. In this case, we say that U is an upper bound for E. We say that E is bounded below if there exists a real number L so that x ≥ L for all x ∈ E.

In this case, we say that L is a lower bound for E. A subset E of R is said to be bounded if E is both bounded above and bounded below.

Let E be a subset of R.

(1) Suppose E is bounded above. An upper bound U of set E is the least upper bounded of E if for any upper bound U⁰ of E, U⁰ ≥ U. If U is the least upper bound of E, we denote U by sup E. The least upper bound for E is also called supremum of E.

(2) Suppose E is bounded below. A lower bounded L of E is said the greatest lower bound of E if for any lower bound L⁰ of E, L ≥ L⁰. If L is the greatest lower bound for E, we denote L by inf E. The greatest lower bound for E is also called the infimum of E.

Example 1.1. Let a, b be real numbers. The set [a, b] = {x ∈ R : a ≤ x ≤ b} is bounded with sup[a, b] = b and inf[a, b] = a.

Example 1.2. Let a be a real numbers. We denote (−∞, a) = {x ∈ R : x < a}. Then (−∞, a) is bounded above but not bounded below.

Example 1.3. Given a sequence (an) of real numbers, let {an∈ R : n ≥ 1} be the image of (a_n), i.e. the set of all values of (a_n). Then (a_n) is bounded (bounded above, bounded below) if and only if the set {a_n∈ R : n ≥ 1} is bounded (bounded above, bounded below).

Theorem 1.1. (Property of R) In R, the following hold:

(1) Least upper bound property: Let S be a nonempty set in R that has an upper bound. Then S has a least upper bound.

(2) Greatest lower bound property: Let P be a nonempty subset in R that has a lower bound. Then P has a greatest lower bound in R.

Example 1.4. Consider the set S = {x ∈ R : x²+ x < 3}. Find sup S and inf S.

Example 1.5. Let S = {x ∈ Q : x² < 2}. Find sup S and inf S.

Example 1.6. Let S = {x ∈ R : x³ < 1}. Find sup S. Is S bounded below?

Proposition 1.1. Let E be a bounded subset of R and U ∈ R is an upper bound of E.

Then U is the least upper bound of E if and only if for any  > 0, there exists x ∈ E so that x ≥ U − .

Proof. Suppose U = sup E. Then for any  > 0, U −  < U. Hence U −  is not an upper bound of E. Claim: there exists x ∈ E so that x > U − . If there is no x so that x > U − ,

then x ≤ U −  for all x ∈ E. This implies that U −  is again an upper bound for E. By the definition of the least upper bound, U ≤ U −  which is absurd since  > 0.

Conversely, let U⁰ = sup E. Since U is an upper bound of E, U ≥ U⁰. Claim U⁰ = U. For any  > 0, choose x ∈ E so that x > U − . Thus U⁰ ≥ x > U − . We know that for any U⁰ > U −  for all  > 0. Since  is arbitrary U⁰≥ U. We conclude that U⁰= U.

 Proposition 1.2. Let E be a bounded subset of R and L ∈ R is a lower bound of E. Then L is the greatest upper bound of E if and only if for any  > 0, there exists x ∈ E so that x ≤ L − .

Proof. We leave it to the reader as an exercise. 

Let E be a nonempty subset of R. We say that M is a maximum of E if M is an element of E and x ≤ M for all x ∈ E. Using the definition, we immediately know that there is only one maximum of E if the maximum elements of E exists. In this case, we denote M by max E. It also follows from the definition that if the maximum of E exists, it must be bounded above.

We say that m is the minimum of a nonempty subset E of R if m is an element of E and x ≥ m for all x ∈ m. We denote m by min E. It follows from the definition that if a set has minimum, it must be bounded below.

Example 1.7. The following subsets of R are all bounded. Hence their greatest lower bound and their least upper bound exist. Determine whether their maximum or minimum exist.

(1) E₁ = (0, 1).

(2) E2 = (0, 1].

(3) E₃ = [0, 1).

(4) E4 = [0, 1].

Proposition 1.3. Suppose E is a nonempty subset of R. If E is bounded above, then the maximum of E exists if and only if sup E ∈ E. Similarly, if E is bounded below, then the minimum of E exists if and only if inf E ∈ E.

Proof. The proof follows from the definition. 

Proposition 1.4. Let E be a nonempty subset of R. Suppose E is a bounded set. Then inf E ≤ sup E.

Proof. We leave it to the reader as an exercise.

 Proposition 1.5. Let E and F be nonempty subsets of R. Suppose that E ⊂ F.

(1) If E and F are both bounded below, then inf F ≤ inf E.

(2) If E and F are both bounded above, then sup E ≤ sup F.

Proof. Let us prove (a). (b) is left to the reader.

For any x ∈ F, x ≥ inf F. Since E is a subset of F, x ≥ inf F holds for all x ∈ E.

Therefore inf F is a lower bound for E. Since inf E is the greatest lower bound for E, inf E ≥ inf F.

 Theorem 1.2. Every bounded monotone sequence is convergent.

Proof. Without loss of generality, we may assume that (an) is a bounded nondecreasing sequence of real numbers. Let a = sup{a_n : n ≥ 1}. Given  > 0, there exists a_N so that a ≥ aN > a − . Since (an) is nondecreasing, an≥ a_N for every n ≥ N. Hence an> a −  for every n ≥ N. In this case, |a_n− a| = a − a_n<  for n ≥ N. We prove that lim

n→∞a_n= a.