Chapter 2

Question 1

Definition: convergence

Answer 1

A sequence (a_n) is said to converge to a limit l∈ℝ if there exists an N ∈Naturals st for all n>N We have |a_n -l | < ε

Question 2

Theorem 2.11: convergence of a sequence to a limit and another limit value

Answer 2

If a sequence converged to a limit then that limit is unique. More precisely if (a_n) if a sequence st lim a_n =l and lim a_n= l’ also (as n tends to Infinity) then l = l’

Proof:
Assume that l is not equal to l’.
Then by definition of convergence we have an M and an N for each limit WE USE ε/2 . By the triangle inequality,
|l-l’| = | l-l’ -a_n + a_n |
Less than or equal to
|a_n-l’| + |l-a_n| =. |a_n-l’| + |a_n-l|
Less than
ε/2 + ε/2 = ε
Since the limits aren’t equal 0 < |l-l’| < ε , holds for any ε. Taking ε = |l-l’| /2 gives the contradiction we require.

Question 3

Definition: divergence

Answer 3

A sequence which fails to converge is said to diverge

(a_n) diverges to infinity if given any k>0 there exists an N in the naturals st for all n >N a_n > k

Respectively a_n < -k for negative infinity

Question 4

Definition: bounded above, bounded, bounded below

Answer 4

A sequence (a_n) is said to be
Bounded above/bounded below/ bounded
If the set {a_n,n∈N } is
Bounded above/ bounded below/ bounded

Question 5

THM: 2.12: convergence and boundedness of sequences

Answer 5

If a sequence (a_n) is convergent then it’s also bounded
PROOF:
1) we need to find a k >0 st |a_n| is less than or equal to K for all naturals. We know (a_n) converges to some l in the reals ….

By triangle inequality
|a_n| less than or equal to |a_n-l | + |l|
Less than ε + |l|
So FOR n>N we take k= ε +|l|
FOR n less than or equal to N: we have |a_n| less than or equal to max({ |a₁|, |a₂|,…,|a_n|})
Hence for all n in the naturals k= max({ |a₁|, |a₂|,…,|a_n|,ε +|l| })

Question 6

THM 2.21 algebra of limits

Proof from book (too long)

Answer 6

Suppose that (a_n) and (b_n) are convergence sequences with the limits l and m respectively.

1) sequences whose nth term is a_n + b_n converges to l+m
2) sequence whose nth term is a_n • b_n converges to lm
3) if c is real then the sequence whose nth term is c•a_n converges to cl
4) if b_n ≉ 0 for all n and also m ≉0 then the sequence whose nth term is a_n / b_n converges to l/m

PROOF

LOOK IN NOTES AS TOO LONG

1) Given any ε>0 there exists an N∈N st if n>N, then |a_n -l | < ε/2 and there also exists N∈N st if n>M |b_n -m| < ε/2. Choose any n > max{N,M} and by the triangle inequality |a_n + b_n - (l +m)|= |a_n -l + b_n -m | less than or equal to | a_n -l| + |b_n -m| < ε/2 + ε/2= ε
2) similarly for n > max{N,M} |a_nb_n -lm|= |a_nb_n- lm +lb_n -lb_n| by triangle inequality is less than or equal to |b_n(a_n -l)| + | l( b_n -m) |= |b_n||a_n-l| + |l||b_n-m|

Assuming l not equal to 0, don’t bother with l=0.

Now as sequence (b_n) is convergent and thus by theorem 2.1.2 is bounded. Hence there is a real number K bigger than 0 st |b_n| ≤ K for all n. So ..

|a_nb_n - on | ≤ K|a_n - l | + |l||b_n - m| …

For any ε> 0 there exists M in N st |b_n -m| < ε/ 2|l| whenever n is bigger than M.

Combining we get for n bigger than Max{M,N},
|a_nb_n - lm| < K ε/2K + |l| ε/2|l| = ε/2 + ε/2 = ε
Probing in case where l not equal to 0 if equal 0 …

3) follows 2) where (b_n) is the constant sequence with nth term c
4) firstly we show that

Question 7

THM 2.2.2: The sandwich rule

Answer 7

Suppose given three sequences (a_n), (b_n) and (c_n), so that for all n∈N we have a_n ≤b_n ≤ c_n. If (a_n) and (c_n) both converge to the same limit l, then (b_n) also converges to l.

Proof:
We must show that given any ε> 0 there exists an N∈N st for all n>N.
|b_n -l | < ε. Ie both b_n -l less than ε and l- b_n less than ε and There exists an M_1 ∈N st for all n bigger than M_1, b_n -n ≤ c_n -l ≤ |c_n -l| < ε. And also there exists an M_2∈N st got all n >M_2 l- b_n≤ l -a_n ≤ |l - a_n| = |a_n -l| < ε.
Taking N =max{M_1,M_2} the result follows.

Question 8

Definition: monotonic increasing and decreasing

Answer 8

A sequence (a_n) is monotonic increasing if a_n+1 ≥ a_n for all n∈N.

It’s monotoníc decreasing if a_n+1 ≤ a_n for all n∈N

We can see that (a_n) is monotonic decreasing if and only if (-a_n) is monotoníc decreasing

Question 9

THM 2.3.1

Monotone convergence theorem

Answer 9

1) if a sequence (a_n) Is bounded above and monotonic increasing then it is convergence and limit as n tends to infinity of a_n = sup_(n∈N)[ (a_n)]
2) if (a_n) is bounded below and monotonía decreasing then convergent and limit as n tends to infinity of a_n = inf_(n∈N)[ (a_n)]

Proof: since (a_n) is bounded above α= sup_(n∈N)(a_n) exists by the completeness property.
By proposition, given any ε>0 there exists an N∈N st a_N bigger than α-ε. As (a_n) is monotonic increasing: for all n bigger than N a_n ≥ a_(N+1) ≥ a_N bigger than α-ε, it follows that |α-a_n| = α -a_n

Similarly 2

Question 10

Monotonía and converges or diverges to

Answer 10

Collorary: if the sequence (a_n) is monotonic increasing / monotonic decreasing then either it converges or diverges to +infinity/ -infinity

Proof:
Suppose (a_n) is monotonic decreasing, then either bounded below or it isn’t. If it is bounded below then it converges by previous theorem and if it isn’t then given any k>0 we can find N in the naturals such that a_n > K, o/w would have lower bound. Since monontonic decreasing have a_n < K for all nbigger than or equal to N sequence diverges to minus infinity

Question 11

A sequence and subsequence

Answer 11

A sequence (y_n) is a subsequence of a sequence (x_n) if there is a strictly increasing function σ: N -> N st y_n = x_σ(n) for all n in the naturals.

Equivalently a subsequence of (x_n, n in N) is a sequence of the form (x_r, r in N) ie (x_n_1 , x_n_2, x_n_3,…) where n_1 < n_2 < n_3

Question 12

Proposition convergence of the subsequence

Answer 12

If the sequence (a_n) converges to some limit l, then every subsequence of (a_n) also converges to l.

Proof:
Given ε>0 , there exists an N∈N st if n bigger than N, then |a_n -l | less than ε.

Let (a_n_r) be a subsequence of (a_n) and choose k∈N st n_k bigger than N. Then for all r bigger than K, |a_n_r - l| less than ε.

Question 13

Monotone subsequence

Answer 13

THM 2.4.2: every sequence has a monotone subsequence

Proof:

Let (a_n) be a sequence and define set
C= {N∈N, a_m < a_N for all m>N}.
The set C is either bounded above or it isn’t. Suppose that C is bounded above: then C is finite.

Choose an n₁∈N as follows:
If C is the empty set then n₁ =1, if C is non empty then n₁= max(C) +1. In either case n₁ is not an element of C for all n bigger than or equal to n₁.

But
•n₁ not in C implies a_n₂ ≥ a_n₁ for some n₂ bigger than n₁
•n₂ not in C implies a_n₃ ≥ a_n₂ for some n₃ bigger than n₂

And then we use induction to completed the proof that (a_n_j ∈N) is monotonic increasing.

If C is unbounded then we can find an infinite sequence (n₁,n₂,n₃,…) in C with n₁less than n₂less than n₃…
•n₁∈C implies a_m less than a_n₁ for all m bigger than n₁ which implies a_n₂ is less than a_n₁
•n₂∈C implies a_m less than a_n₂ for all m bigger than n₂ which implies a_n₃ is less than a_n₂
And again we use induction to complete the proof that (a_n_k, k∈N) us monotonic decreasing.

Question 14

Boleando- Weirstrass

Answer 14

Every bounded sequence has a convergent subsequence.

Proof:
Suppose (a_n) is bounded. By theorem 2.4.2 it has a monotone subsequence, (a_n_r), which is itself bounded.

Then (a_n_r) converges by theorem that states monotone sequence converges if bounded. (Thm2.3.1)

Question 15

Cauchy sequence

Answer 15

A sequence (a_n) is said to be a Cauchy sequence if given any ε> 0 , there exists an N ∈N st for all m and n > N we have | a_m - a_n| < ε

Every convergent sequence is Cauchy
Every Cauchy sequence is bounded

Arbitrarily close terms but says nothing in convergence.

Question 16

THM Cauchy sequence and converges …

Answer 16

If (a_n) is a Cauchy sequence in R, then it converges to a limit in R

Proof: by every convergent sequence being Cauchy

We have that (a_n) is bounded and hence by the Bolzano-Weirstrass theorem is has a subsequence (a_n_r) converging to some l in the reals. Proving that (a_n) converges to l:

Given any ε>0 there exists an M₁∈N st for all m,n bigger than M₁ we have |a_m -a_n| < ε/2 and there exists M₂ ∈N st n_r bigger than M₂.
So we have that |a_n_r -l| < ε/2.

Hence for all n bigger than M₁ and choosing r so that n_r is bigger than max{M₁,M₂} we have
|a_n -l| ≤ |a_n -a_n_r| + |a_n_r - l| < ε/2 + ε/2 = ε

Question 17

Definition: complete subset and Cauchy

Answer 17

A non-empty subset A subset of R is said to be complete if every Cauchy Sequence taking values in A converges to a limit in A.

Ie R is complete by theorem for (a_n) Cauchy sequence in R always converges to a limit in R

Q is not complete eg approximating root 2. By rationals

Question 18

Convergent sequences and Cauchy …

Answer 18

Every convergent sequence is Cauchy:

Proof
Problem 41

Question 19

Cauchy and bounded …

Answer 19

Every Cauchy sequence is bounded:

Proof:
Problem 42

Question 20

(a_n) and a_n

Answer 20

(a_n) is sequence and nth term is a_n

Question 21

What else modulus tells us in the convergence of a sequence |a_n - l | < ε

Answer 21

|a_n - l | < ε if and only if l-ε < a_n < l + ε

We interpret Into equality and so we can show it converges if we show the rhs

Convergence also tells us that as n tends to infinity |a_n - l | gets very small

Question 22

Proof of convergence given sequence:

Answer 22

We find a SPECIFIC N for ε, that is we found N from ε and thus taking this N, for all n above, can show that in fact |a_n - l | < ε

Question 23

Notation vs definition of convergence

Answer 23

Notation is not the definition.

Notation would be lim as n tends to infinity of a_n =l etc not to do with |a_n - l | < ε

Question 24

Example: prove that lim as n tends to infinity of r^n =0 whenever 0 ≤ r < 1

Answer 24

BY CASES:
•r=0: clearly for all n∈N r^n =0
• 0< r < 1: we have that 1/r > 1.
Let 1/r = 1 + h where h >0.

Then by BERNOUILLS INEQUALITY we have
1/r^n = ( 1 +h )^n ≥ 1 +nh
And thus r^n ≤ 1/ (1+nh).

We must show that, given any ε>0 there exists an N∈N st for all n > N |r^n| = r^n < ε. So we require an N st (1+Nh)^-1 less than ε which implies that N bigger than (1-ε) / εh by algebra.

By the ARCHIMEDEAN PROPERTY for real (1-ε) / εh there does exists such an N. Then for all n>N. r^n ≤ (1+mg)^-1 < 1/ (1+Nh) < ε

Question 25

Example- find limit as n tends to infinity of (2n -7n^3)/(6n^2 +11n^3)

Answer 25

We divide both by the highest power in the denominator: n^3
Hence it becomes the limit as n tends to infinity of

2/n^2 -7 divided by 6/n + 11n

Which becomes -7/11

Question 26

Example: find the limit as n tends to infinity of cos(n)/n

Answer 26

We use the sandwich rule. As for all n in the naturals cos(n is bounded bewueen -1 and 1. Then using the sandwich rule with -1/n and 1/n the limit is 0

Question 27

Example- EXAM STYLE QUESTION

Consider the sequence (a_n) given by
a_1=0, a_(n+1) = (3a_n +1)/(a_n +3)
For all n bigger than 1.

• Use induction to show that a_n is bounded within [0,1] for all n in the naturals
• Show that (a_n) is monotonic increasing
• Explain why limit as n tends to infinity of a_n exists and find its value

Answer 27

• true for n=1 base case, inductive hyp gives a_n+1 bigger than 0
  a_n+1 -1 shown to be less than or equal to 0 also. By Induction etc
• for all n in the naturals we can show that a_n - a_n+1 is less than or equal to 0. (Numerator of a fraction is)
• by previous parts it converges by theorem 2.3.1 convergence of monotoníc bounded

By taking limits on both sides of recurrence relation limit as n tends to unfinished is 1

Question 28

Example- The golden section as a limit Constructing by recursion
a_1=1 and a_(n+1) = sqrt ( 1+ a_n)

Answer 28

Calculating first few terms (6) looks like increasing and bounded above.
Proving this: looks like 2 is an upper bound. Prove this by contradiction. Suppose there exists an N such that a_n is less than or equal to 2 for all n less than or equal to N and that a_(N+1) is bigger than 2. From our calculations we know it’s at least bigger than 6. Substitutions in the recurrence in a_(N+1) is bigger than 2 and squaring gives that 1+ a_N is bigger than 4. Which gives a_N is bigger than 3. But this is a contradiction. increaskng. Is proven by squaring the recursion and subtracting equations for a_n and a_n-1 is show it’s bigger than 0. Must combine all equations to cancel and give numbers on denominator.

To how u has a limit use theorem and take limits of the recurrence relation and use algebra of limits. Two sols are found.

Question 29

Example : e as a limit
e = limit as n tends to infinity of (1+ 1/n) to the power of n

Proving that the limit exists and that e is bounded between 2 and 3

Answer 29

We show minotnix increasing and bounded above .
Monotmonic increasing:
By the theorem of means, and using a_1= a_2=..a_n-1 = 1+ (1/n-1) and a_n =1.

Geometric mean (1+ (1/(n-1)))^((n-1)/(n))

Arithmetic mean is 1+ 1/n
Hence the theorem of means raised to the power of n gives equations for a_n and a_n+1 in inequality to show increasing.
To show bounded use the BINOMIAL theorem to expand and inequalities from this to show bounded above by using a geometric series compared

Question 30

Guess the limit as n tends to infinity of the sequence whose nth term is

3/4 • ( 0.5 - (7/n^3)) and prove

Answer 30

Limit is 3/8:

We need to show that given an epsilon bigger than 0 we can find an N in the naturals so that if n is bigger than N
| a_n -l | = | -21/4n^3 | = 21/4n^3 is less than epsilon.

This occurs if and only if n is bigger than (21/4epsilon)^(1/3). BY THE ARCHIMIDEAN PROPERTY such an N exists and so for n bigger than N
21/4n^3 is less than 21/4N^3 is less than epsilon

And thus the limit is 3/8.