Analysis period 1.2

Question 1

Sequence

Answer 1

A sequence is a real-valued function whose domain is a set of the form {n∈Z| n>=m} when m is a fixed integer

Question 2

Convergence

Answer 2

A sequence {an} coverges to A ∈R <=> ∀ epsilon >0 there exists n* ∈N : |an - A| < epsilon ∀n >= n*. Here A is the limit of the sequence {an}

Question 3

Proving convergence

Answer 3

1. Always start with let epsilon > 0 be given
2. Start with |an - A| and get rid of the absolute values as much as you can.
3. Next, you can make this term smaller than any other, bigger term. If you can get the latter term smaller than epsilon, then the term you began with is also smaller than epsilon. Note: It could be that your inequality is only valid from a certain point. Take this point into account when finding your n*
4. Let n* be the maximum of the values fir which your inequalities hold and the function of epsilon
5. Mention that there exist n∈N s.t. |an - A| < epsilon for all n >= n and draw your conclusion.

Question 4

Can two covergent sequences have the same limits?

Answer 4

Any two limits of a convergent sequence are the same. This means that a convergent sequence has a unique limit

Question 5

When is a sequence said to be bounded?

Answer 5

A sequence {an} is said to be bounded <=> there exist M>0: |an| <= M for all n∈N.

If a sequence is not bounded, then it is said to be unbounded

Question 6

Is a convergent sequence bounded?

Answer 6

Yes

Take note that an bounded sequence does not have to be convergent.

Question 7

Sequences unequal to 0

Answer 7

Consider a sequence {an} that converges to A =/ 0
Then there exists n∈ N: an =/ 0 for all n >= n. In fact, |an| >= |A|/2 for all n >= n*.

Question 8

Special sequence

Answer 8

If a real number, r, satisfies |r| < 1, then lim r^n = 0

Question 9

Absolute value of the sequence

Answer 9

The sequence {an} coverges to 0 if and only if the sequence {|an|} converges to 0

Question 10

Squeeze theorem

Answer 10

Suppose that {an}, {bn} and {cn} are three sequences and suppose that there exists n1 s.t. an <= bn <= cn for all n >= n1. If {an} and {cn} both converge to A, then {bn} must also converge to A.

Question 11

Product of sequence

Answer 11

If sequence {an} converges to 0 and sequence {bn} is bounded, then {anbn} converges to 0.

Question 12

Proof product of sequence

Answer 12

Let epsilon > 0 be given. We want to have that |anbn - 0| < epsilon. To do so, we have to find nstar.
|anbn - 0| = |anbn| = |an| x |bn|
|anbn| <= M x |an|
there exists nstar ∈N: |an - 0| = |an| < epsilon / M
there exists nstar ∈N: |anbn - 0| <= M x |an| < M x epsilon/ M = epsilon

So the sequence {anbn} converges to 0

Question 13

Diverging sequence

Answer 13

A sequence {an} diverges to infinity if and only if for all M>0 there exist nstar ∈ N: an > M for all n >= nstar.

If this is the case, we say that lim (n to inf) an = infinity
If -an > M for all n >= nstar then lim ( n to inf) an = - infinity so we have divergence to - infinity

Question 14

ut

Comparison theorem

Answer 14

If a sequence {an} diverges to infinity and there exist n1 ∈ N s.t. an <= bn for all n >= n1, then the sequence {bn} must also diverge to infinity

Question 15

Proving divergence to +- infinity

Answer 15

1. Always start with Let M>0 be given
2. Start with (-)an
3. Next you can find an inequality with this term to some other, smaller term. If you can find a way to get the latter term bigger than M, the term you began with is bigger than M as well. Note: It could be that your inequality is only valid form a certain point. Take this point into account when finding nstar.
4. Let nstar be the maximum of values for which your inequalities are valid and the function of M.
5. Stay that there exist nstar ∈ N s.t.: (-)an > M for all n>= nsatr and draw your conclusion.

Question 16

Properties diverging sequences

Answer 16

If a sequence {an} diverges to infinity and a sequence {bn} is bounded below by K, then
1. {an + bn} diverges to infinity
2. {anbn} diverges to infinity, provided that K>0
3. {c An} diverges to infinity for all c>0
4. {c an} diverges to - infinity for all c<0

Question 17

Inverse diverging sequences

Answer 17

Consider a sequence {an} where an>0 for all n∈N.
Then {an} diverges to infinity if and only if the sequence {1/an} converges to 0.

Question 18

Ratio test

There are three options

Consider a sequence {an} with an =/ 0 for all n∈N s.t. lim (n to inf) |an+1 / an| = alpha, alpa ∈ R

Answer 18

1. If alpha<1, then lim(n to inf) an = 0 (an converges to 0)
2. If alpha >, then lim(n to inf)|an| = infinity (an diverges)
3. If alpha = 1, then {an} may converge, diverge to +/ inf or oscillate

Question 19

Monotonic convergence theorem

For increasing and decreasing

Answer 19

For an increasing sequence {an}, there are tw possibilities
1. {an} is bounded above by a constant M, in which case there exist L<=M: lim(n to inf) an = L
2. {an} is unbounded in which case lim(n to inf) an = inf

For a decreasing sequence {an}, there are two possibiities:
1. {an} is bounded below by a constant M, in which case there exists L>=M: lim(n to inf) an = L
2. {an} is unbounded, in which case lim(n to inf) an = -inf

Question 20

Neighborhood

Answer 20

Let epsilon>0 be given.
A neighborhood for a real number s is set in the set: Neps(s) = Neps = {x∈R | |x - s| < epsilon}.
A Deleted neighborhood of s is a set: Neps(s)- = Neps- = {x∈R | 0<|x - s|< epsilon}.
Observe that a neighborhood of s is an interval (s - epsilon, s + epsilon) for any epsilon>0.
A deleted neighborhood of s is a neighborhood of s without the central value s.

Question 21

Accumulation point

Answer 21

Let S be a set of real numbers. A real number s0 is an accumulation point of S if and only if for all epsilon > 0, there exist t∈S: 0<|t - s0|< epsilon.
In other words, s0 is an accumulation point if and only if for all epsilon>0 there exists t∈S: s0 ∈ Neps-(t)

Every bounded, infinite set of real numbers has at least one accumulation point.

Question 22

Cauchy sequence

Answer 22

A sequence {an} is called cauchy if and only if for all epsilon>0 there exists nstar∈N: |am - an| < epsilon, for all m,n >= nstar

Every cauchy sequence is bounded. Not every bounded sequence is cauchy.
In R, a sequence is cauchy if and only if it is convergent.

Question 23

Proving sequence to be cauchy

Answer 23

1.Let epsilon>0 be given
2.Start with |am - an| and get rid of the absolute values
3.Next, you can make this term smaller than any other, bigger term. If you can get the latter term smaller than epsilon, then the term you began with is also smaller than epsilon. Note: it coulld be that your inequality is only valid upward of a certain point. Take this point into account when finding your nstar.
4.Let nstar be the maximum of the values for which your inequalities hold and the function of epsilon.
5. Say that ther exist nstar∈N s.t |am - an|< epsilon for all n,m>= nstar and draw your conclusion.

Question 24

Contractive sequence

Answer 24

A sequence {an} is said to be a contractive sequence if and only if there exist k∈(0,1) |an+2 - an+1|<= K|an+1 - an| for all n∈N.

Every contractive sequence is a cauchy sequence, and hence, convergent.

Question 25

Subsequence

Answer 25

The sequence {bn} is a subsequence of {an} if and only if every value of {bn} is contained in {an}

Question 26

Subsequential limit

Answer 26

A is a subsequential limit point of a sequence {an} if and only if there exists a subsequence of {an} that converges to A.

Question 27

Subsequence for bounded sequences

Answer 27

Any bounded sequence must have at least one convergent subsequence.

Question 28

Subsequence for convergent sequences

Answer 28

A sequence converges to A if and only if each of its subsequences converges to A.

Question 29

Upper limit and lower limit

Answer 29

Let T be the set of all subsequential limits of {an}. Sup T is called the limit superior (upper limit) of {an} and we can write
sup T = lim (n to inf) sup an = lim(n to inf) an
If infinity ∈ T then sup T = infinity

Inf T is called the limit inferior (lower limit) of {an} and we write
inf T = lim (n to inf) inf an = lim(n to inf) an
If - infinity ∈ T, then inf T = - infinity