4 Sequences

Question 1

Definition 4.1:
Sequence

Answer 1

A sequence of real numbers is a function from N to R.

Question 2

Definition 4.2:
Convergent sequence

Answer 2

A real sequence (aₙ)n∈N converges to L∈R if for every ε > 0 there exists N∈N such that |aₙ − L| < ε for all n ≥ N.
L is the limit of the sequence.
We denote convergence of (aₙ)n∈N to L by:
* aₙ→L as n→∞
* L = lim_(n→∞) aₙ.

(∀ ε > 0) (∃ N∈N) s.t. ∀ n ≥ N, |aₙ − L| < ε

If a sequence does not converge, then we say that is diverges.

Question 3

Proposition 4.1:
Convergence of
1. Constant sequence (c, c, c, …) where c∈R
2. (1/n)n∈N
3. (aⁿ)n∈N where a∈R,|a|<1

Answer 3

The sequence converges to:
1. c
2. 0
3. 0

Question 4

Proposition 4.2:
Relationship between limits of two sequences

Answer 4

Suppose that (aₙ)n∈N and (bₙ)n∈N are real sequences with aₙ ≤ bₙ for all n ∈ N. If
* L = lim.(n→∞) aₙ and
* M = lim.(n→∞) bₙ

exist, then L ≤ M.

Question 5

Corollary 4.1:
Relation when aₙ is convergent and aₙ [≤, ≥] b

Answer 5

Suppose that (aₙ)n∈N is a convergent sequence.
* If b∈R with aₙ ≤ b for all n∈N, then lim.(n→∞) aₙ ≤ b.
* If c∈R with aₙ ≥ c for all n∈N, then lim.(n→∞) aₙ ≥ c.

Question 6

Theorem 4.1:
Uniqueness of limits

Answer 6

A sequence of real numbers has at most one limit.

If a sequence converges, then that limit is unique.

Question 7

Definition 4.3:
Bounded sequence

Answer 7

A real sequence (aₙ)n∈N is bounded if the set of all its members S = { a₁, a₂, … } is bounded.

Question 8

Proposition 4.3:
Relationship between covergent and bounded sequences

Answer 8

Any convergent sequence is bounded.

Question 9

Theorem 4.2:
Algebra of limits

Answer 9

Let (aₙ)n∈N and (bₙ)n∈N be convergent sequences with limits A = lim.n→∞(aₙ) and B = lim.n→∞(bₙ). Then for any constant C ∈ R,

1. lim.n→∞(an + bn) = A + B,
2. lim.n→∞(an − bn) = A − B,
3. lim.n→∞(Can) = CA,
4. lim.n→∞(anbn) = AB,

and if bn≠0 for all n∈N and if B≠0, then
5. lim.n→∞(an/bn) = A/B.

Question 10

Proposition 4.4:
Relation between an and |an|

Answer 10

Let (an)n∈N be a real sequence.
1. [1] If an →L as n→∞ for some L∈R, then |an|→|L| as n→∞.
2. If |an|→0 as n→∞, then an →0 as n→∞ as well.

Question 11

Definition 4.4:
Divergent sequence

Answer 11

• (an)n∈N diverges to ∞ if for all M ∈ R there exists N∈N
  such that an > M for all n≥N. i.e. an →∞ as n→∞ or limn→∞(an) = ∞.
• (an)n∈N diverges to −∞ if for all M ∈ R there exists N∈N such that an < M for all n∈N. i.e. an →−∞ as n→∞ or limn→∞(an) = −∞.

Question 12

Definition 4.5:
increasing, decreasing and monotone sequences

Answer 12

A real sequence (an)n∈N is
1. increasing if an ≤ an+1 for all n ∈ N,
2. strictly increasing if an < an+1 for all n ∈ N,
3. decreasing if an ≥ an+1 for all n ∈ N,
4. strictly decreasing if an > an+1 for all n ∈ N.

Question 13

Theorem 4.3:
Relationship between bounded monotone sequences and convergence

Answer 13

If a real sequence is monotone and bounded, then it converges.

Question 14

Theorem 4.4:
Sandwich theorem

Answer 14

Suppose that (an)n∈N, (bn)n∈N, and (cn)n∈N are real sequences with an ≤ bn ≤ cn for all n ∈ N.
If lim.n→∞ an = limn→∞ cn = L, then (bn) also converges and limn→∞ bn = L.

Question 15

Proposition 4.5:
Property of (1/an)n∈N if limn→∞ an = 0

Answer 15

If (an)n∈N is a real sequence such that limn→∞ an = 0 and an ̸= 0 for all n ∈ N, then (1/an)n∈N is divergent.

Question 16

Lemma 4.1:
an ≤ Csⁿ

Answer 16

Suppose that (an)n∈N is a sequence of positive numbers such that r = limn→∞ a(n+1) / an exists.
Then for any s > r there exists a number C > 0 such that an ≤ Csⁿ for all n ∈ N.

Question 17

Theorem 4.5:
Growth factor test

Answer 17

Let (an)n∈N be a sequence of positive numbers.
r= limn→∞ a(n+1) / an with 0 ≤ r ≤ ∞.
1. If 0 ≤ r < 1, then an→0 as n→∞.
2. If r > 1, then an→∞ as n→∞.
3. If r = 1, then the test is inconclusive.

Question 18

Corollary 4.3:
Limits of sequences involving polynomials and exponentials

Answer 18

“Exponential trumps polynomial”
Let k ∈ N and a > 1. Then as n → ∞:
* nᵏ/aⁿ → 0
* aⁿ/nᵏ → ∞

Question 19

Definition 4.6:
Subsequence

Answer 19

Suppose that (an)n∈N is a sequence of real numbers and (nk)k∈N is a strictly increasing sequence of natural numbers. Then (ank )k∈N is called a subsequence of (an)n∈N.

Question 20

Proposition 4.6:
Property of subsequences if the sequence converges.

Answer 20

A sequence converges to L if all its subsequences also converge to L

Question 21

Corollary 4.4:
A sequence diverges if its subsequences…

Answer 21

If two subsequences of (an) converge to different limits, then (an) is divergent

Question 22

Theorem 4.6:
Bolzano-Weierstrass

Answer 22

Every bounded real sequence contains a convergent subsequence.

Question 23

Definition 4.7:
Cauchy sequence

Answer 23

∀ε > 0, ∃N ∈ N such that ∀m,n ≥ N |am − an| < ε.

It is a Cauchy sequence if the terms of the sequence eventually all become arbitrarily close to one another.

Question 24

Theorem 4.7:
Relation between convergent and Cauchy sequence

Answer 24

A sequence (an)n∈N converges if, and only if, it is a Cauchy sequence.

Question 25

Proposition 4.7:
sup.n≥k[an] and inf.n≥k[an]

Answer 25

1. If (an)n∈N is bounded above, then the sequence (sup.n≥k[an])k∈N is a decreasing real sequence. Otherwise, sup.n≥k[an] = ∞ for all k∈N.
2. If (an)n∈N is bounded below, then the sequence (inf.n≥k[an])k∈N is an increasing real sequence. Otherwise, inf.n≥k[an] = −∞ for all k∈N.

Question 26

Definition 4.8:
limsup.n→∞(an)

Answer 26

Let (an)n∈N be a real sequence. If it is bounded above, then
limsup.n→∞(an) = lim.k→∞( sup.n≥k[an] ).
Otherwise, limsup.n→∞(an) = ∞.

Question 27

Definition 4.8:
liminf.n→∞(an)

Answer 27

Let (an)n∈N be a real sequence. If it is bounded below, then
liminf.n→∞(an) = lim.k→∞( inf.n≥k[an] ).
Otherwise, liminf.n→∞(an) = −∞.

Question 28

Proposition 4.8:
Relation between liminf and limsup

Answer 28

For any real sequence (an)n∈N,
liminf an ≤ limsup an.

Question 29

Theorem 4.8:
Convergent sequence and its relation with limsup and liminf

Answer 29

A real sequence (an)n∈N converges if, and only if, limsup.n→∞(an) = liminf.n→∞(an) = L∈R
Then:
lim.n→∞(an) = limsup.n→∞(an) = liminf.n→∞(an).