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Page 1 of 13

Sequences (notes by S. Sands) [Chapter 2 in Lebl]

Section 16: Convergence * denotes a very important theorem!

Definitions.

A sequence is a function whose domain is the set N of natural numbers.

Notation: (sn) or (s1, s2, s3, ...) or {sn} denotes a sequence. {sn} notation used in Lebl book.

sn is called the nth term of the sequence.

If we want a different starting subscript, we may write �������� = (sm, sm+1, sm+2, ...)

Examples:

(a) sn = 1 + (−1) n (0, 2, 0, 2, 0, 2, ...)

(b) (2, 4, 6, ...) = (sn), where sn = 2n.

(c) sn = 1 + 1/2 n

(3/2 , 5/4, 9/8, 17/16, ...)

The first graph shows the sequence graphed as a function f(n) = 1 + 1/2 n

for n in N.

The second graph shows the sequence values plotted on the number line.

Page 2 of 13

Lebl definition is on page 44.

Definition. A sequence (sn) is said to converge to the real number s provided that

for each ε > 0 there exists a real number N such that for all n ∈ N, n > N implies that | sn - s | < ε .

If (sn) converges to s, then s is called the limit of the sequence (sn), and

we write lim�→� �� = � , or lim �� = � or sn→ s.

If a sequence does not converge to a real number, it is said to diverge.

Example (sn) = (1/n)

It appears that lim (1/n) = 0.

Suppose ε = 0.3

If we go far enough out in the sequence, we want |1/n - 0| < 0.3

Note that 1/n < 0.3 implies that n > 1/0.3 ≈ 3.3

So, if we choose any positive integer n > 3, we have |1/n - 0| < 0.3

Prove that lim (1/n) = 0.

Discussion:

Suppose ε is any positive real number.

We want to find N such that if n > N, |1/n - 0| < ε.

So, we want 1/n < ε, which implies that n > 1/ε.

Therefore:

Formal Proof:

Given any ε > 0, let N = 1/ε.

Then n > N implies that |1/n - 0| = 1/n < 1/N = ε.

Example: Prove that lim (1/√�) = 0. Discussion:

Suppose ε is any positive real number.

We want to find N such that if n > N, |1/√� - 0| < ε. So, we want 1/√� < ε, which implies that √� > 1/ε and so n > (1/ε)2 = 1/ε2 Therefore:

Formal Proof:

Given any ε > 0, let N = 1/ε 2

Then n > N implies that |1/√� - 0| = 1/√� < 1/√� = 1/�1/�� = ε.

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Example: Prove that lim (1 + 1/2 n ) = 1

Discussion:

Suppose ε is any positive real number.

We want to find N such that if n > N, |(1 + 1/2 n )

- 1| < ε.

So, we want 1/2 n < ε. We could solve for n.

But we could use an easier observation: 1/2 n < 1/n for all positive integers n.

So if we choose n so that 1/n < ε, then n > 1/ε.

Formal Proof:

Given any ε > 0, let N = 1/ε.

Then n > N implies that |(1 + 1/2 n )

- 1| = 1/2

n < 1/n < 1/N = ε.

Example: Prove that lim �������� = � �.

Discussion:

Suppose ε is any positive real number.

We want to find N such that if n > N, ��������� − � �� < �

����������������������� � < �

� � �������� < �

� ������� < � . Solve for n to get � � ! +

� � < �.

We will choose N = � � ! +

� �.

Formal Proof:

Given any ε > 0, let N = � � ! +

� �

Then n > N implies that � > � � ! + � �

7� > � �! + 4

7� − 4 > � �!

and we have

��������� − � �� = �

� �������� = �

� � ∙

� ����� < �

� � ∙

�! � � = ε as desired.

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*Theorem. Let (sn) and (an) be sequences of real numbers and let s in R. If for some k > 0 and some m ∈ N we have

| sn - s | ≤ k| an |, for all n > m, and if lim an = 0, then it follows that lim (sn) = s.

Proof:

Suppose ε is any positive real number.

We want to find N such that if n > N,| sn - s | < ε.

But | sn - s | ≤ k| an | for all n > m and now we want k| an | < ε.

Since lim an = 0, there exists real number N1 such that n > N1 implies | an | < ε/k.

So, let N = max{m, N1}.

Therefore, n > N implies that | sn - s | ≤ k| an | < k(ε/k) = ε.

*Example lim n1/n = 1.

Suppose ε is any positive real number.

We want to find N such that if n > N, | n 1/n

- 1 | < ε.

Since n 1/n

≥ 1 for all positive integers n, the quantity bn = n 1/n

- 1 is nonnegative.

Then n 1/n

= 1 + bn.

so n = (1 + bn) n . By the binomial theorem (pages 106-107, can be proven by induction), for n ≥ 2,

� = 1 + �'� + �� ��� − 1�'�� + ⋯ + '��

≥ 1 + �� ��� − 1�'��

So,

� − 1 ≥ �� ��� − 1�'��

� � ≥ '��

)�� ≥ '� Therefore n ≥ 2 implies that | n

1/n - 1 | = bn ≤ )�� = √2 ∙

� √�

But lim (1/√�) = 0, as shown earlier, and now by applying the preceding theorem, we must have lim n

1/n = 1.

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A sequence (sn) is bounded if the range { sn : n ∈ N } is a bounded set. [Lebl defined on p. 43]

That is, (sn) is bounded if there exists real number M such that | sn |≤ M for all n ∈ N.

*Theorem. Every convergent sequence is bounded. [Prop. 2.1.7, Lebl, page 45] Proof:

Suppose sequence (sn) converges to s.

Applying the limit definition with ε = 1, there exists N such that

n > N implies that | sn - s | < 1.

Therefore, n > N implies that | sn | = | sn - s + s |

≤ | sn - s | + | s | by the triangle inequality

< 1 + | s |

So, let M = max{ |s1|, | s2|, ..., |sN| , 1 + | s |}

Then | sn | ≤ M for all n ∈ N, so (sn) is bounded.

Theorem. If a sequence converges, its limit is unique. [Prop. 2.1.6, Lebl, page 45]

Proof:

Suppose that a sequence (sn) converges to both s and t. We will show that s = t.

Since lim sn = s, given any ε > 0, there exists real number N1 such that

n > N1 implies | sn - s | < ε/2

Since lim sn = t, given any ε > 0, there exists real number N2 such that

n > N2 implies | sn - t | < ε/2

Let N = max{N1, N2}.

Then n > N implies that |s - t | = | s - sn + sn - t | adding and subtracting sn

≤ | s - sn | + | sn - t | by the triangle inequality

< ε/2 + ε/2

= ε

So, given any ε > 0, |s - t | < ε.

By (previously assigned homework on ordered fields), we can conclude that | s - t | = 0.

But this means s - t = 0, so s = t as claimed.

Example: sn = 1 + (−1) n diverges. As seen earlier, this sequence is (0, 2, 0, 2, 0, 2, ...).

No matter how large an N we pick, there are values of n, m > N for which sn =0 and sm =2. The sequence

values do not get close to any one value, and by our previous theorem, a convergent sequence can have

only one limit. Therefore, this sequence cannot converge, and so it diverges.

This example illustrates that a bounded sequence is not necessarily convergent.

Page 6 of 13

Section 17: Limit Theorems * denotes a very important theorem!

*Theorem. [Section 2.2.2, Lebl, page 53] Suppose that (sn) and (tn) are convergent sequences with lim (sn) = s and lim (tn) = t. Then

(a) lim (sn + tn) = s + t.

Sum theorem: The limit of the sum is the sum of the limits.

(b) lim (k sn) = ks and lim (k + sn) = k + s, for any k in R.

(c) lim (sn tn) = st.

Product theorem: The limit of the product is the product of the limits.

(d) lim (sn / tn) = s/t, provided that tn ≠ 0 for all n and t ≠ 0.

Quotient theorem: The limit of the quotient is the quotient of the limits, provided the limit of the

denominator is nonzero.

Discussion of (a):

Suppose lim (sn) = s and lim (tn) = t.

Suppose ε is any positive real number.

We want to find N such that if n > N, | (sn + tn) - (s + t)| < ε.

But | (sn + tn) - (s + t)| = | (sn - s) + (tn - t) | ≤ | sn - s | + | tn - t | and we can make | sn - s | and| tn - t|

as small as we please.

Formal Proof of (a):

Suppose lim (sn) = s and lim (sn) = t.

Given any ε > 0,

there exists N1 such that n > N1 implies | sn - s | < ε/2 and

there exists N2 such that n > N2 implies | tn - t | < ε/2.

Let N = max{N1, N2}.

Then n > N implies that | (sn + tn) - (s + t)|

= | (sn - s) + (tn - t) |

≤ | sn - s | + | tn - t | by the Triangle Inequality

< ε/2 + ε/2

= ε

Therefore, lim (sn + tn) = s + t.

Discussion of (c):

Suppose lim (sn) = s and lim (tn) = t.

Suppose ε is any positive real number.

We want to find N such that if n > N, | sn tn - st| < ε.

But | sn tn - st | = | sn tn - sn t + sn t - st | ≤ | sn tn - sn t | + | sn t - st |

= | sn || tn - t | + | t | | sn - s | ≤ M| tn - t | + | t | | sn - s |

since the convergent sequence (sn) must be bounded, and now we can make | sn - s | and| tn - t| as

small as we please.

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Formal Proof of (c):

Suppose lim (sn) = s and lim (tn) = t.

Suppose ε is any positive real number.

Since (sn) converges, (sn) is bounded. So, there exists M such that | sn | ≤ M for all n in N.

Let K = max {M, |t|}.

Given any ε > 0,

there exists N1 such that n > N1 implies | sn - s | < ε/(2K) and

there exists N2 such that n > N2 implies | tn - t | < ε/(2K).

Let N = max{N1, N2}.

Then n > N implies that | sn tn - st | = | sn tn - sn t + sn t - st |

≤ | sn tn - sn t | + | sn t - st | by the triangle inequality

= | sn || tn - t | + | t | | sn - s |

≤ M| tn - t | + | t | | sn - s | since (sn) is bounded

≤ K| tn - t | + K | sn - s | by definition of K

< ε/(2K) + ε/(2K)

= ε

Therefore, lim (sn tn) = st.

Discussion of (d):

Suppose lim (sn) = s and lim (tn) = t.

lim (sn / tn) = lim (sn ⋅(1/ tn) ) = lim (sn ) ⋅ lim (1/ tn) by (c) provided that lim (1/ tn) exists

= s (1/t) = s/t, provided that lim (1/ tn) is equal to 1/t.

So, it suffices to show that lim (1/ tn) = 1/t.

By hypothesis, tn ≠ 0 for all n and t ≠ 0.

Suppose ε is any positive real number.

We want to find N such that if n > N, | 1/tn - 1/t| < ε.

But | 1/tn - 1/t| = | (t - tn)/(tnt)|= | 1/( tnt)| | tn - t | . If we can get an upper bound on | 1/( tnt)|, then

we can make| tn - t| as small as we please and we will have what we want. To get an upper bound on

|1/( tn)|, we want a lower bound on | tn |, and we can use the fact that when n is large, tn is close to t,

and by hypothesis that t is nonzero.

| tn | = | t - (t - tn)| ≥ |t| - | t - tn |

If we take n large enough, then | t - tn | < |t|/2,

so -| t - tn | > -|t|/2, and

|t| - |t - tn| >|t| - |t|/2 = |t|/2

Then | 1/( tnt)| | tn - t | = |1/t| ⋅ |1/tn | | tn - t | < 1/|t|⋅ 2/|t| | tn - t |= 2/t 2 | tn - t |.

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Formal Proof of lim (1/ tn) = 1/t, provided tn ≠≠≠≠ 0 for all n and t ≠≠≠≠ 0.

Suppose lim (tn) = t, where tn ≠ 0 for all n and t ≠ 0.

Suppose ε is any positive real number.

Since (tn) converges to t and t is nonzero,

there exists N1 such that n > N1 implies | tn - t | < |t|/2 and

there exists N2 such that n > N2 implies | tn - t | < t 2 /2 ε.

Let N = max {N1, N2 }.

Then n > N implies that | 1/tn - 1/t| = | (t - tn)/(tnt)|

= | 1/( tnt)| | tn - t |

= |1/t| |1/ tn | | tn - t |

< [1/|t| 2/|t| | tn - t |

= 2/t 2 | tn - t |

< 2/t 2 (t

2 /2 ε)

= ε

Therefore, lim (1/ tn) = 1/t.

Example: Prove that lim �������� = � �. In the previous section, we proved the limit using the definition.

Now we can use the limit theorem.

lim �������� = lim ��+, ��-,

dividing numerator and denominator by n

= ./0 1�� + ,2

./0 1��-,2 applying the quotient theorem of limits (d)

= ./0 �� ./0 + ,

./0 �� ./0 -, applying the sum theorem of limits (a)

= ./0 �� ./0 + ,

./0�� � ./0 +, applying part (b) of the limit theorem

= ��3����3� = � � applying the facts that lim c = c (part b of theorem) and lim (1/n) = 0.

Page 9 of 13

Theorem. [Lemma 2.2.3, p. 52 Lebl]

Suppose that (sn) and (tn) are convergent sequences with lim (sn) = s and lim (tn) = t.

If sn ≤ tn for all n in N, then s ≤ t.

Proof (by contradiction).

Suppose not. Suppose that lim (sn) = s, lim (tn) = t, sn ≤ tn for all n in N, but s > t.

Then s - t > 0.

Let ε = (s - t)/2, so ε > 0 and t = s - 2ε.

Since lim (sn) = s, there exists N1 such that n > N1 implies | sn - s | < ε.

so -ε < sn - s < ε, i.e., s - ε < sn < s + ε.

Similarly, since lim (tn) = t, there exists N2 such that n > N2 implies | tn - t | < ε, so t - ε < tn < t + ε

Let N = max {N1, N2 }.

Then n > N implies that

tn < t + ε = (s - 2ε) + ε = s - ε < sn which contradicts the assumption that sn ≤ tn.

Corollary. If (tn) converges to t and (tn) ≥ 0 for all n in N, then t ≥ 0. [Corollary 2.2.4, p. 53 Lebl]

*SQUEEZE Lemma [Lemma 2.2.1, p. 51 Lebl] Suppose that (sn) and (tn) are convergent sequences with lim (sn) = lim (tn).

If sn ≤ xn ≤ tn for all n in N, then (xn) converges and lim (sn) = lim (xn) = lim (tn).

*Theorem. (Ratio test for sequences converging to 0). [Variation of Lemma 2.2.12(i), Lebl, page 59]

Suppose (sn) is a sequence of positive terms and that the sequence of ratios ( sn + 1 / sn ) converges to L.

If L < 1, then lim (sn) = 0.

Proof:

Since L < 1, there exists a real number c such that L < c < 1.

Let ε = c - L, so ε > 0.

Since ( sn + 1 / sn ) converges to L, there exists N such that n > N implies that | sn + 1 / sn - L | < ε.

So, sn + 1 / sn < L + ε = L + (c - L) = c.

We have sn + 1 < c sn for all n > N.

Let K = smallest integer > N.

Then 0 < sn < c sn-1 < c 2 sn-2 < ... < c

n-K sK for all n > K.

c n-K

sK = s K c

-K c

n = M c

n for constant M = s

K c

-K

So we have 0 < sn < M c n for all n > K.

lim M c n = 0 because 0 < c < 1 (earlier homework exercise)

Therefore we have a positive sequence (sn) bounded above by a sequence that converges to 0,

so lim sn = 0 by our theorem in the previous section.

Page 10 of 13

Example: (sn) = (5 n /n!) converges to 0.

sn + 1 / sn = 4,5+ �����! ÷

4, �! =

4,5+ �����∙�! ∙

�! 4, =

4 ���.

Then lim ( sn + 1 / sn ) = lim 4

��� = 0 < 1, so lim (5 n /n!) = 0.

*Theorem. (Ratio test for diverging sequences). [Variation of Lemma 2.2.12(ii), Lebl, page 59]

Suppose (sn) is a sequence of positive terms and that the sequence of ratios ( sn + 1 / sn ) converges to L.

If L > 1, then (sn) is unbounded and diverges.

Example: (sn) = (5 n /n!) diverges.

*Example: Let x > 0. [Prop. 2.2.11, Lebl, page 58]

(i) If x < 1, then lim x n = 0

(ii) If x > 1, then (x n ) is unbounded and diverges.

These results can easily be proven by applying the Ratio Tests for sequences.

------------------------------------------------------

Infinite Limits

Definition. A sequence (sn) is said to diverge to +∞∞∞∞ and we write lim (sn) = +∞ provided that

for every M in R there exists a number N such that n > N implies that sn > M.

Similarly, (sn) is said to diverge to -∞ and we write lim (sn) = -∞, provided that

for every M in R there exists a number N such that n > N implies that sn < M.

Theorem. Suppose (sn) and (tn) are sequences such that sn ≤ tn for all n in N.

(a) If lim (sn) = +∞, then lim (tn) = +∞.

(b) If lim (tn) = -∞, then lim (sn) = -∞.

*Theorem. Let (sn) be a sequence of positive numbers. Then lim (sn) = +∞ iff lim (1/sn) = 0.

Page 11 of 13

Section 18: Monotone Sequences and Cauchy Sequences * denotes a very important theorem!

[Section 2.1.1 in Lebl]

Definition. A sequence (sn) of real numbers is increasing if sn ≤ sn + 1 for all n in N.

It is decreasing if sn ≥ sn + 1 for all n in N.

A sequence is monotone if it is increasing or decreasing.

Examples:

(sn) = (1/n) is monotone decreasing.

(sn) = (2n) is monotone increasing. (sn) = (2 - 1/n) is monotone increasing.

(sn) = ((-1) n /n) is not monotone.

*Theorem. (Monotone Convergence Theorem)

A monotone sequence is convergent if and only if it is bounded. [Th 2.1.10, Lebl, page 46]

Proof:

Suppose that (sn) is a convergent monotone sequence. Then it must be bounded (since any convergent

sequence is bounded.)

Suppose that (sn) is a bounded monotone sequence. (For convenience, assume that the sequence is

increasing. The proof for the decreasing case is similar.) We want to show that (sn) converges.

Let S = {sn : n in N}. Since (sn) is bounded, S is bounded. Clearly, S is nonempty.

Since S is nonempty and bounded above, by the completeness axiom, S has a supremum.

Call it s = sup S. Since s is an upper bound for S, sn ≤ s for all n in N.

Claim: lim sn = s.

Given any ε > 0, s - ε cannot be an upper bound for S.

So there exists N such that sN > s - ε.

Since (sn) is increasing, then n > N implies that sn ≥ sN > s - ε .

Now we have s - ε < sn ≤ s. So, -ε < sn - s ≤ 0 and we have | sn - s | < ε. Therefore, lim sn = s.

Page 12 of 13

Recursively defined sequences - Section 2.2.3 Lebl

Example: Let (sn) be defined recursively as s1 = 1 and ���� = �1 + �� for n ≥ 1. �� = 1 �� = √2 ≈ 1.414 �� = �1 + √2 ≈ 1.554 �� = )1 + �1 + √2 ≈ 1.598 We want to establish that (sn) is a bounded increasing sequence. [Then by the Monotone Convergence

Theorem, (sn) converges.]

Claim (1): (sn) is bounded above by 2.

Proof (by induction):

For n = 1, s1 = 1 < 2.

Suppose that sk < 2 for some k in N.

Then �=�� = �1 + �= < √1 + 2 = √3 < 2. Therefore, by induction, sn < 2 for all n in N.

Claim (2): (sn) is an increasing sequence. That is, for all n in N, �� ≤ ����. Proof (by induction):

For n = 1, �� = 1 < √2 = ��.

Suppose that �= ≤ �=��for some k in N.

Then �=�� = �1 + �= ≤ �1 + �=�� = �=��.

Therefore, by induction, �� ≤ ���� for all n in N.

By the Monotone Convergence Theorem, we know that (sn) converges to some number s.

Now we want to find s.

Since lim ���� = lim �� lim �1 + �� = lim ��

and we have √1 + � = �.

But this means 1 + s = s 2 , so s

2 - s - 1 = 0 and solving the quadratic equation, we get � = �±√4� .

We have seen that sn ≥ 1 for all n, so the limit s cannot be ��√4 � .

Therefore, the sequence converges to � = ��√4� .

This is a number with a storied history. It is called the golden ratio. For more, see

http://mathworld.wolfram.com/GoldenRatio.html

Page 13 of 13

Section 19: Subsequences [Section 2.1.3 in Lebl]

Definition. Let (sn) be a sequence. Let (nk) be a sequence of natural numbers such that n1 < n2 < n3 < ....

The sequence A��BC is called a subsequence of (sn).

That is, we take a sequence and remove some (possibly infinitely many) terms, keeping the order of the

remaining terms intact. If we have infinitely many terms remaining, then what is left is a subsequence.

Note that n1 ≥ 1, n2 ≥ 2, n3 ≥ 3,..., and in general, nk ≥ k.

Example: (sn) = 1��2 = 11, � � ,

� � ,

� � ,

� 4 ,… 2

Some subsequences: 1 �� , � � ,

� F ,

� G ,… 2 ,1

� � ,

� ,

� �� ,

� G� ,… 2. Not a subsequence: 1

� � ,

� � ,

� F ,

� � ,

� 4 ,

� � … 2 (out of order)

Theorem. [Prop. 2.1.17, p. 49 Lebl]

If a sequence (sn) converges to a real number s, then every subsequence of (sn) also converges to s.

Proof:

Let A��BC be any subsequence of (sn) . Let ε > 0.

Since (sn) converges, there exists N such that n > N implies that | sn - s | < ε.

But nk ≥ k. So, when k > N, we have nk > N, which implies that | ��B - s | < ε. Thus, A��BC converges to s.

Corollary: If two subsequences of (sn) converge to different numbers, then the sequence (sn) diverges.

*Example: Prove that lim x 1/n

= 1 for any real number x with 0 < x < 1.

Proof:

Suppose 0 < x < 1. Then 0 < x 1/n

< 1 for all n, so (x 1/n

) is a bounded sequence.

x 1/(n + 1)

- x 1/n

= x 1/(n + 1)

( 1 - x 1/n - 1/(n+1)

) = x 1/(n + 1)

( 1 - x 1/[n(n + 1)]

) > 0 for all n.

So x 1/(n + 1)

- x 1/n

> 0, which implies x 1/(n + 1)

> x 1/n

for all n, and so (x 1/n

) is an increasing sequence.

By the Monotone Convergence Theorem, (x 1/n

) converges. Call the limit s.

Note that (x 1/n

) 2 = x

1/(2n) , and (x

1/(2n) ) is a subsequence of (x

1/n ), so (x

1/(2n) ) also converges to s.

Therefore, lim (x 1/n

) 2 = lim x

1/(2n)

(lim x 1/n

) 2 = lim x

1/(2n)

s 2 = s.

So, s 2 - s = 0

s(s - 1) = 0

s = 0 or s = 1. But s1 = x > 0 and (x 1/n

) is increasing, so the limit s cannot be 0. Therefore s = 1.

We conclude that lim x 1/n

= 1.

Example: Use the corollary to show that sn = (-1) n = (-1, 1, -1, 1, ...) diverges.

The subsequence is 1, 1, 1,... converges to 1, and another subsequence -1, -1, -1,... converges to -1.

Since there are subsequences with different limits, (sn) cannot converge.