Sequences: Maths 1st Partial

0.0(0)
Studied by 0 people
call kaiCall Kai
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/54

encourage image

There's no tags or description

Looks like no tags are added yet.

Last updated 2:21 PM on 10/9/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai
Chat

No analytics yet

Send a link to your students to track their progress

55 Terms

1
New cards

What is a sequence

An ordered list of up to infinitely many numbers where f: ℕ+ → ℝ

2
New cards

What are two features of a sequence as a set

  • Composed of only isolated points

  • Not open


3
New cards

am can also be written as

am = f(m)

4
New cards

Whats the formula for a sequence being bounded above

∃ k such that am < k ∀ m

5
New cards

What’s the formula for a sequence being bounded

∃ k such that |am| < k ∀ m

6
New cards

We say that {am} satisfies a property ___ if there is an integer m̄ such that a satisfies the property ___

We say that {am} satisfies a property eventually if there is an integer m̄ such that a satisfies the property ∀ m > m̄

the property is eventually satisfied if, past a certain location in the sequence, every point satisfies the property

7
New cards

How do we define monotone for a sequence?

am > am+1 ∀ m

8
New cards


Mathematical definition for how we say: {aₘ} is convergent to L ∈ ℝ if:

∀ ε > 0, ∃ mε such that m ≥ mε ⇒ |aₘ − L| < ε

there exists a number greater than zero such that for every number in the set am past a certain position mε, am is so close to L that am - L = this tiny number greater than 0, and therefore am is a sequence converging to L — often looks like the ‘tired frog’ sequence

9
New cards

Alternative neighbourhood based method:

We say that {aₘ} is convergent to L ∈ ℝ if:

∀ Bε(L), ∃ mε such that m ≥ mε ⇒ aₘ ∈ Bε(L)


for every small neighbourhood radius sigma around centre L, as you reach a certain point in the sequence, all numbers past that point in the sequence are an element of this tiny neighbourhood around the limit

10
New cards

We say that {aₘ} is positively divergent if:

∀ K, ∃ mₖ such that m ≥ mₖ ⇒ aₘ > K

for every large number k, there exists a certain location in the sequence where, past this point, every number in the sequence is greater than K

11
New cards

Neighbourhood-based definition of

We say that {aₘ} is positively divergent if:

∀ B(+∞), ∃ mₖ such that m ≥ mₖ ⇒ aₘ ∈ B(+∞)

For every neighbourhood of positive infinity (extending from K to +∞), there exists a position on the sequence past which every number is an element of this neighbourhood.

12
New cards

Give an example of a limit from above and how thats denoted

am = 1 / m

lim am = 0


Lim am = L+

13
New cards

What does ∀ Bε+(L) refer to

[L, L + ε)

14
New cards

How do we mathematically define an am limited from above

∀ Bε+(L), ∃ mε such that m ≥ mε ⇒ aₘ ∈ Bε⁺(L)

( or equivalently:

aₘ ∈ [L, L + ε)

or:

L ≤ aₘ < L + ε )

15
New cards

Give an example of an am with a limit from below

am = 1 - 1/m

a frog that forever leaps closer and closer from 0 to 1 but never quite gets to 1

16
New cards

What is am when lim am is:

  • L

  • +∞

  • -∞

  • None


L — am convergent

+∞ — am positively divergent

-∞ — am negatively divergent

none — am oscillating or irregular


am is regular in the first three cases - when convergent or divergent

17
New cards

Give an example of osciallating bounded and oscillating unbounded

bounded: Sin(m)

unbounded: (-1)m m2

18
New cards

What is implied by am being regular

It has a limit

19
New cards

What’s the logic behind lim am = +∞ ⇔ lim 1/am = 0+

am infinite ⇔ 1 / am infinitesimal

1 / 0+ = +∞

1 / +∞ = 0+

The reciprocal of a divergent sequence is equal to zero

20
New cards

What is called an infinite sequence and what is called an infinitesimal sequence

Infinite: lim am = +∞

Infinitesimal: lim bm = 0

21
New cards

What does 0+ represent

0+ represents a limit where a positive variable approaches zero from above, and the limit evaluates to 0

22
New cards

am has a limit in R extended ⇒

the limit is unique

23
New cards

Convergent am ⇒

Convergent am ⇒bounded am

24
New cards

Monotone am ⇒

Monotone am ⇒ regular am

25
New cards

If am is monotone bounded ⇒

am monotone bounded ⇒ am convergent

lim am = L- (approaching from below)

26
New cards

If am increasing unbounded ⇒

am increasing unbounded ⇒ am positively divergent

lim am = L

27
New cards

If am decreasing unbounded ⇒

am decreasing unbounded ⇒ am negatively divergent

28
New cards

What is lim ma if a:

a > 0

a = 0

a < 0

lim ma when a:

a > 0 +∞

a = 0 1

a < 0 0+

<p>lim m<sup>a</sup> when a:</p><p>a &gt; 0        +<span>∞</span></p><p>a = 0        1</p><p>a &lt; 0        0<sup>+</sup></p>
29
New cards

What is the limit of a geometric sequence lim qm when q is:


q > 1

q = 1

-1 < q < 1

q = -1

q < -1

lim of qm:

q > 1 +∞

q = 1 1

-1 < q < 1 0

q = -1 irregular bounded

q < -1 irregular unbounded

30
New cards

what is lim logam when a:

a > 1

0 < a < 1

a > 1 +∞

0 < a < 1 -∞

<p>a &gt; 1          +∞</p><p>0 &lt; a &lt; 1    -∞</p>
31
New cards

What is lim (ln m)a when alpha:

a > 0

a = 0

a < 0

a > 0 +∞

a = 0 1

a < 0 0+

32
New cards

lim am = A, lim bm = B (provided not indeterminate cases e.g. infinity minus infinity

What is lim (am + bm)

lim (am + bm) = A + B

33
New cards

lim (am • bm)

lim (am • bm) = A • B

34
New cards

lim (am / bm)

lim (am / bm) = A / B

35
New cards

How do we transform 00 into regular notation using ab = eloga ^b = eb loga

00 transforms into e0(-∞)

36
New cards

How do we transform +∞0 into regular notation using ab = eloga ^b = eb loga

+∞0 = e0(+∞)

37
New cards

How do we transform 1∞ into regular notation using ab = eloga ^b = eb loga

turns into e∞ ln1

38
New cards

What is the comparison criterion

Let a < bm < cm be such that at least eventually

lim am = lim cm = L

Then lim bm = L

39
New cards

What is the ratio criterion

If lim | am+1 / am | = q with q < 1

Then lim am = 0


so to do this u take ur formula for am and substitute in (am + 1) then u just divide these two things by each other and mess around with the nummbers to find its limit and see if its less than 1, if it equals 1 then the test is inconclusive

40
New cards

What does lim (lower in the scale / higher in the scale) equal?

. lower in the scale

lim ————————— = 0

. higher in the scale

41
New cards

What is the order among infinities (when am, bm → ∞) for lim (am / bm)


knowt flashcard image
42
New cards

How do ln m, ma, and qm relate to each other on a vague graph

knowt flashcard image
43
New cards

What is the limit by comparing infinitesimals am,bm → 0 for lim (am / bm)

knowt flashcard image
44
New cards

What do we call am when it is a “faster” infinitesimal than bm

am is an infinitesimal of higher order than bm

45
New cards

the exponential is a ___ limit than any power

the exponential is a faster limit than any power

46
New cards

When you have multiple infinitesimals to deal with, you first keep ___

the lower infinitesimals. the faster infinitesimals become negligible (e.g. 0.1 + 0.0000001 ~ 0.1)

47
New cards

lim ___ = ea

lim (1 + a / m )m = ea

48
New cards

lim am / bm = 1 means am ____ to bm

am asymptotic to bm

49
New cards

lim am / bm = 0 means am ____ to bm

lim am / bm = 0 means am negligible with respect to bm

50
New cards

when am ~ bm , am, bm are ____

when am ~ bm , am, bm are asymptotic

am and bm are ‘quite’ equal

51
New cards

when am ~ bm , lim(am / bm) =

when am ~ bm , lim(am / bm) = 1

52
New cards

when am = o(bm), am ____ to bm

when am = o(bm) little o bm, am negligible with respect to bm

53
New cards

when am = o(bm) , lim(am / bm) =

when am = o(bm) , lim(am / bm) = 0

54
New cards

What is a mathematical way of writing

am ~ bm ⇔

am ~ bm ⇔ am = bm + o(bm)

55
New cards

What is the formal mathematical definition for convergence of a sequence of vectors

lim xₘ = L

∀ ε > 0, ∃ mε such that m ≥ mε ⇒ d(xₘ, L) < ε

or equivalently:

∀ ε > 0, ∃ mε such that m ≥ mε ⇒ ‖xₘ − L‖ < ε


there exists a sigma where, past a certain point in the sequence amε , the distance between the vector points of the sequence and L is sigma, where sigma is close to zero