A sequence (an) is convergent to a real number a if for any given  &gt

A sequence of real numbers is function a from the set of positive integers N to the set of real numbers R. If n ∈ N, we denote a(n) by an. We denote a sequence a : N → R by (an).

In this note, a sequence means a sequence of real numbers.

Definition 1.1. A sequence (a_n) is convergent to a real number a if for any given  > 0, there exists N ∈ N such that

|a_n− a| < , whenever n ≥ N_. In this case, a is called a limit of (an).

Definition 1.2. A sequence (an) is said to be convergent if (an) has a limit. Otherwise, (a_n) is said to be divergent.

Proposition 1.1. Suppose (an) is convergent. Let a and b be its limit. Then a = b. In other words, the limit of a convergent sequence is unique.

Proof. The proof will be given later. 

By Proposition of (1.1), we know if a limit of a sequence exists, it is unique. Hence if (a_n) is convergent to a, we denote a by lim

n→∞a_n.

Let x be a real number. We define [x] to be the largest integer less than x. Then [x] ≤ x < [x] + 1.

Example 1.1. Show that lim

n→∞

1 n = 0.

Proof. Let  > 0. Choose N_ = [1/] + 1. Then N_> 1/. If n ≥ N_, then 

   1 n− 0

= 1 n ≤ 1

N

< , ∀n ≥ N_.

 Example 1.2. Show that lim

n→∞

1 2ⁿ = 0.

Proof. We can use induction to prove that 2ⁿ > n for any n ≥ 1. This implies that for all n ≥ 1,

1 2ⁿ − 0

= 1 2ⁿ < 1

n.

For any  > 0, we choose N = [1/] + 1. Then for any n ≥ N, n > 1/. This gives us that for any n ≥ N_,

1 2ⁿ − 0

< 1 n < .

 Example 1.3. Show that

n→∞lim

n²− n + 5 3n²+ n + 7 = 1

3.

1

Proof. Observe that for n ≥ 5,

|x_n−1

3| ≤ n + 5 3n² < 2

3n.

Solve _3n² < . We obtain n > _3². Take N= [2/(3)] + 1. Then for n ≥ N,

|x_n−1 3| < .

 Example 1.4. Show that lim

n→∞2ⁿ¹ = 1.

Proof. Let us use the formula xⁿ− 1 = (x − 1)(xⁿ⁻¹+ · · · + x + 1) and choose x = 2^1/n. Then

2ⁿ¹ − 1 = 1

2ⁿ⁻¹ⁿ + · · · + 2ⁿ¹ + 1 . Since 2 > 1, 2^1/n > 1. Then

|2ⁿ¹ − 1| < 1 n. Given  > 0, take N_= [1/] + 1. Then for any n ≥ N_,

|2¹ⁿ − 1| < .

 Let us go back to the proof of uniqueness of limits. Suppose (an) is convergent to both of a and b, given  > 0, we can choose N_⁰, N_⁰⁰∈ N so that |an− a| < /2 whenever n ≥ N_⁰ and |an− b| < /2 whenever n ≥ N_⁰⁰. Let N = max{N_⁰, N_⁰⁰}. For n ≥ N_, |an− a| < /2 and |bn− b| < /2. By triangle inequality,

|a − b| ≤ |a_n− a| + |a_n− b| <  2+ 

2 = .

This shows that |a − b| <  for any  > 0. This happens only when a = b.

2. Monotone Sequence Properties Let us define a sequence (xn) as follows:

(2.1) xn= 1 + 1

1!+ 1

2!+ · · · + 1 n!.

How do we know that the sequence is convergent or not? Observe that for each n ≥ 1, xn+1= xn+ 1

(n + 1)! =⇒ xn+1> xn. Hence (x_n) is a nondecreasing sequence.

Definition 2.1. A sequence (an) is called

(1) nondecreasing if an≤ a_n+1 for all n; (when an< an+1, we say that the sequence is increasing)

(2) nonincreasing if an≥ a_n+1 for all n; (when an> an+1, we say that the sequence is decreasing)

(3) monotone if it is either nondecreasing or nonincreasing.

Since we know that n! > n(n − 1) for any n ≥ 2, xn= 2 + 1

2!+ 1 3! + 1

4!+ · · · + 1

(n − 1)! + 1 n!

< 2 + 1

1 · 2 + 1

2 · 3 + · · · + 1

(n − 2) · (n − 1)+ 1 (n − 1) · n

= 2 +

 1 −1

2

 + 1

2 −1 3



+ · · · +

 1

n − 2− 1 n − 1

 +

 1

n − 1 − 1 n



= 3 − 1 n.

Then x_n< 3 for any n ≥ 1, i.e. (x_n) is bounded above.

Definition 2.2. A sequence (an) is said to be

(a) bounded above if there exists U ∈ R so that an≤ U for any n;

(b) bounded below if there exists L ∈ R so that an≥ L for any n;

(c) bounded if it is bounded above and bounded below, i.e. L ≤ an≤ U for any n.

Theorem 2.1. (Monotone Sequence Property) Every bounded monotone sequence is con- vergent.

3. Properties of Limits

Lemma 3.1. A convergent sequence is always bounded.

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



Theorem 3.1. Let (an) and (bn) be two convergent sequences whose limit are a and b respectively. 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 (an) and (bn) 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+ bn− (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

anbn− ab = a_nbn− ab_n+ abn− ab = (a_n− a)b_n+ a(bn− b).

Assume that a 6= 0. Since (b_n) 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 (bn) 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 3.1. Suppose (a_n) 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 3.2. Let (an) and (bn) be two convergent sequence of real numbers. Suppose that a_n≤ b_n for all n ≥ 1 and lim

n→∞a_n= a and lim

n→∞b_n= b. Then a ≤ b.

Proof. Since both (a_n) and (b_n) 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 −  < a_n< b_n< b + .

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



Proposition 3.1. Suppose (an) is convergent to a. Assume an≥ 0 for all n. For k ≥ 1,

n→∞lim

√k

a_n=√^k a.

Proof. Using Lemma 3.2, we know that a ≥ 0 since an≥ 0 for all n ≥ 1.

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

|√^k

a_n| < , whenever n ≥ N_. This shows that lim

n→∞

√k

a_n= 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 an≥ 0,

 ^k

√an−√^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

√a_n−√^k a

< |a_n− a|

a^k−1k

< a^k−1k  a^k−1k

= .

This proves that lim

n→∞

√k

an= √^k a.



Example 3.1. Let (an) and (bn) and (cn) be sequence of real numbers. Suppose

n→∞lim a_n= 1, lim

n→∞b_n= −1, lim

n→∞c_n= 3.

Evaluate the following limits.

(1) lim

n→∞(2an− 3b_n).

(2) lim

n→∞anc³_n. (3) lim

n→∞

√cn. (4) lim

n→∞

anbn

c_n .

Example 3.2. Let (a_n) be the sequence defined by an+1=√

2 + an, n ≥ 1 with a1 =√

2. Evaluate lim

n→∞an using the property of limits.

Example 3.3. Evaluate the limit of each convergent sequence using the properties of limit.

(1) a_n= 3n²− n + 1

5n²+ 6n + 7, for n ≥ 1.

(2) lim

n→∞(√

n + 1 −√ n).

(3) lim

n→∞

(−2)ⁿ+ 3ⁿ (−2)ⁿ⁺¹+ 3ⁿ⁺¹. (4) lim

n→∞

1 + a + · · · + aⁿ

1 + b + · · · + bⁿ, where |a| < 1 and |b| < 1.

(5) lim

n→∞

 1 n² + 2

n² + · · · +n − 1 n²

 .

Example 3.4. Are the statements in Theorem 3.1 still true if one of limn→∞an and limn→∞bn does not exist? If the statements are false, give some counterexamples.

Proposition 3.2. Let (an) be a sequence of real numbers. Suppose

n→∞lim a2n= lim

n→∞a2n+1= a.

Then (an) is convergent to a.

Proof. This will be left to the reader as an exercise. 

Example 3.5. Define a sequence (a_n) of real numbers by a_n+1= 2 + a_n

1 + an

, n ≥ 1 and a1 = 1.

(1) Show that (a_2n−1) is increasing.

(2) Show that (a_2n) is decreasing.

(3) Show that both (a2n) and (a2n−1) are convergent.

(4) Show that (a_n) is convergent and evaluate the limit using the properties of limit.

Exercise.

4. The Euler number e Let (x_n) 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 y_n=

 1 + 1

n

n

, n ≥ 1.

This this section, we are going to prove that (y_n) is convergent and lim

n→∞x_n = lim

n→∞y_n. Before proving the convergence of (y_n), let us recall some basic math facts learnt from high school.

Theorem 4.1. (A.G. inequality) Let a₁, · · · , a_n be any nonnegative real numbers. Then a1+ · · · + an

n ≥ √ⁿ

a₁· · · a_n.

Proof. The proof will be given in the appendix. 

Theorem 4.2. (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)!.

Proof. The proof will be given in the appendix.

 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 +_n¹

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

yn+1=



1 + 1 n + 1

n+1

>

 1 + 1

n

n

= yn.

Hence (yn) 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 x_n< 3 for all n ≥ 1, and hence

y_n≤ x_n< 3

for all n ≥ 1. This shows that (yn) is bounded. We conclude that (yn) is convergent by Theorem 2.1. We let us denote y = lim

n→∞y_n.

Theorem 4.3. Let (x_n) and (y_n) be as above and x and y be their limits respectively.

Then x = y.

To prove this theorem, we need the following lemma.

Corollary 4.1. Let (an) be a convergent sequence whose limit is a. Suppose that an≤ M for all n ≥ 1. Then a ≤ M.

Proof. Let us take b_n= M for all n ≥ 1. Using Lemma 3.2, we see that a ≤ M.

 Since yn≤ x_n for all n ≥ 1, using Lemma 3.2, we see that y ≤ x. On the other hand, for each 1 ≤ k ≤ n,

y_n= 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

y_n> 1 + 1 +n 2

 1

n² + · · · +n k

 1 n^k, we see that by Lemma 3.2,

y ≥ 1 + 1 + 1

2! + · · · + 1 k! = x_k for all k ≥ 1. Using Corollary 4.1, we see that

x = lim

k→∞xk ≤ y.

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

Definition 4.1. Let (x_n) be the sequence defined in (2.1). We denote e = lim

n→∞x_n. The number e is called the Euler number.

We also obtain from the previous discussion that Theorem 4.4.

n→∞lim

 1 + 1

n

n

= e.

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

5. Sandwich Principle

Theorem 5.1. (Sandwich Principle) Let (a_n), (b_n), (c_n) be sequences of numbers. Suppose that there exists N > 0 so that for any n ≥ N , an ≤ b_n ≤ c_n, and (an), (cn) are both convergent to a. Then lim

n→∞b_n= a.

Proof. Exercise. 

Example 5.1. Suppose that the sequence (a_n) satisfies the following inequality n − 1

2n ≤ a_n≤ n + 3 2n . Find the limit of (an).

Since lim

n→∞

n − 1 2n = 1

2 = lim

n→∞

n + 3

2n , by the Sandwich principle, lim

n→∞an= 1 2. Example 5.2. Find the limit of each convergent sequence.

(1) lim

n→∞

√3

n²sin(n!) n + 1 . (2) lim

n→∞

2ⁿ n!. (3) lim

n→∞

√n

2ⁿ+ 3ⁿ. Solution:

(1). For any θ ∈ R, we have −1 ≤ sin θ ≤ 1. Therefore for any n ≥ 1,

−

√3

n² n + 1 ≤

√3

n²sin(n!) n + 1 ≤

√3

n² n + 1. Since lim

n→∞ −

√3

n² n + 1

!

= lim

n→∞

√3

n²

n + 1 = 0, by the Sandwich principle, lim

n→∞

√3

n²sin(n!) n + 1 = 0.

(2) For n ≥ 3, since each 2/j < 1 for j ≥ 3, we have 0 < 2ⁿ

n! = 2 1

  2 2

  2 3



· · ·

 2

n − 1

  2 n



< 2 · 2 n = 4

n. Since lim

n→∞0 = lim

n→∞

4

n = 0, by the Sandwich principle, lim

n→∞

2ⁿ n!. (4) is left to the reader.