Topic 5: Sequences and Series Flashcards

1
Q

What is the definition of a sequence (2)

A

-A sequence is an infinite list of real numbers
-Equivalently, we can view this as function f: ℕ -> ℝ

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2
Q

What is the definition of a sequence converging to a limit (4)

A

-Let (an) be a sequence, and suppose L ∈ ℝ is some finite real number
-Then an tends/converges to the limit L, if for any ε > 0, there exists N ∈ ℕ such that |an - L| < ε for all n > N
-an tends/converges to any Y value if for a very small number ε (a Y value error number), there exists an x value N such that the absolute distance between the sequence and the limit is less than the y error value for all x values greater than N
-We write lim(n -> ∞) an = L or an -> L as n -> ∞

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3
Q

How can we prove an = (2n+3)/(3n-4) -> 2/3 (7)

A

-Let an = (2n+3)/(3n-4), set L = 2/3
-Suppose |an - L| = |(2n+3)/(3n-4) - 2/3| < ε
-|17/(9n-12)| < ε
-For n>1, 17/(9n-12) < ε => n > (17+12ε)/9ε
-so if n > (17+12ε)/pε, |an - 2/3| < ε
-Set N = [(17+12ε/9ε]
-Therefore, for n > N, |an - 2/3| < ε

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4
Q

How can we define a sequence tending to ∞/-∞ (2)

A

-A sequence (an) tends to ∞ if, given any A>0, there exists some N = N(A) ∈ ℕ such that an > A for all n > N (for any Y value A there exists some x value a where at that point you have Y value A, and the y value of the sequence is greater than A for all x values greater than N)
-Similarly, an -> -∞ if for A < 0, there exists N = N(A) ∈ ℕ such that an < A for all n > N

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5
Q

How can we propose one sequence tends to 0 if we know another one does (2)

A

-Let (an) and (bn) be sequences, and suppose there exists some N ∈ ℕ such that an > bn for all n > N.
-Then, if bn -> ∞ as n -> ∞, we know an -> ∞

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6
Q

How can we prove one sequence tends to 0 if we know another one does ()

A

-Let A>0
-Since bn -> ∞, there exists N1 ∈ ℕ such that bn > A for all n > N1
-Now choose N = max {N0, N1}, where N0 is the point where an > bn
-So for n > N we have an > bn > A, so an also -> ∞ as n -> ∞

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7
Q

What is the definition of a convergent vs divergent sequence (2)

A

-A sequence (an) is convergent if it tends to a finite limit as n -> infinity
-A sequence is divergent if it tends to +- infinity as n -> infinity

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8
Q

What is the definition of a bounded sequence (2)

A

-A sequence (an) is bounded if there exists some finite M > 0 such that |an| ≤ M for all n ∈ ℕ
-A sequence is bounded below if there exists M_ ∈ ℝ such that an ≥ M_ for all n ∈ ℕ, and bounded above if opposite

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9
Q

Are all bounded sequences convergent (1)

A

-No, not all bounded sequences are convergent (sin is bounded between +- 1 but doesn’t converge)

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10
Q

How can we prove that convergent sequences are bounded (1,6)

A

-Proposition: convergent sequences are bounded

Proof:
-Suppose (an converges to finite limit L as n -> ∞
-Then there exists some N ∈ ℕ such that for any ε > 0, we have |an - L| < ε for n > N
-In this case, chose ε = 1
-Rearranging this and using the triangle inequality gives us |an| < L + 1 for all n > N
-Now let M = max {|an, |L + 1|}
-Then for all n ∈ ℕ, we have |an| ≤ M

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11
Q

What are 2 examples of the algebra of limits (2)

A

Suppose an -> a and bn -> b as n -> ∞

-(an + bn) -> (a + b) as n -> ∞
-(an)(bn) -> (ab) as n -> ∞

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12
Q

What is the sandwich rule (2)

A

Theorem:
-Suppose an ≤ bn ≤ cn for all n > N ∈ ℕ
-if an -> L, cn -> L as n -> ∞, bn -> L

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13
Q

How can we prove the sandwich rule (FINISH FLASHCARD WITH PAPER ()

A

-Let ε > 0
-Since an -> L, as n -> ∞, there exists some N1 ∈ ℕ such that |an - L| < ε for n > N
-Similarly, there exists some n

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14
Q

What is an increasing/decreasing sequence (2,2)

A

-A sequence is increasing if an+1 ≥ an for all n ∈ ℕ
-A sequence is strictly increasing if an+1 > an for all n ∈ ℕ

-A sequence is decreasing if an+1 ≤ an for all n ∈ ℕ
-A sequence is strictly decreasing if an+1 <
an for all n ∈ ℕ

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15
Q

What is a monotonic sequence (2)

A

-A sequence is monotonic if it is either increasing or decreasing for n ∈ ℕ
-A sequence is strictly monotonic if it is either strictly increasing or decreasing for n ∈ ℕ

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16
Q

What happens to the upper limit of an increasing sequence, and what theorem can we get as a result of this (2,1,1)

A

If a sequence is increasing, it is either
-Bounded above
-Unbounded

-Corollary: an increasing sequence converges if and only if it is bounded above

-Theorem: A bounded monotonic sequence converges

17
Q

What is an example of a subsequence (2)

A

-Suppose an = 1/(2n) = 1/2, 1/4, …
-If we delete every even-numbered entry, we get a’n = 1/(22n-1)

18
Q

What is the definition of a subsequence (1,3)

A

-A subsequence is a sequence that can be derived from another sequence by deleting some/no elements without changing the order of the remaining elements

-Suppose (an)n=1 is a sequence
-Let (nk)k=1 be a strictly increasing sequence of positive integers
-Then (ank)k=1 = an1 > an2 is a subsequence of (an)n=1

19
Q

What is the relationship between the limits of sequences and subsequences (2)

A

-Suppose (an) is a convergent sequence where an -> L as n -> ∞
-Then any subsequence of (an) also converges to L as n -> ∞

20
Q

How do we prove that every sequence has a monotonic subsequence (3, 2, 4,1)

A

-We say element ap of sequence an is a terrace point if ap ≥ an for all n ≥ p (kind of like a max point for a sequence)
-Every point after a terrace point is lower than it
-The sequence either has infinitely many terrace points, or finitely many (or 0)

-If an has infinitely many terrace point, we can label these as an1, an2…, where n1 < n2
-So an1, an2… is a decreasing subsequence of an

-If an has finitely many (or 0) terrace points, find n1 such that all the terrace points ap occur for p < n1
-Since n1 isn’t a terrace point, there must be some n2 > n1 with an2 > an1
-Since n2 also isn’t a terrace point, there must be some n3 > n2 with an3 > an2
-We can then continue in this way to show the existence of infinitely many n1, n2 …. such that an1 < an2 …, which is a strictly increasing subsequence

-Therefore, an contains either a decreasing or strictly increasing subsequence, which is therefore monotonic

21
Q

What is the Bolzano-Weierstrass theorem (1)

A

-Every bounded sequence has a convergent subsequence

22
Q

How can we write the sum of the first n terms of gk = ark-1 (3)

A

-Suppose a, r ∈ ℝ, with r not equal to 1
-Consider (gr)k=1 defined by gk = ark-1
-Sn = ∑gk = a((1-ra)/(1-r))

23
Q

How can we write the sum of the first n terms of gk = ark-1 if |r| < 1 (3)

A

-If |r| < 1, rn -> 0 as n -> ∞
-Thus, ∑ark-1 = a/1-r
-This is the geometric series

24
Q

What is the definition of a series (2)

A

-A series is the sum of the (infinitely many) terms in a sequence
-We write this as ∑k=1 an, given sequence an

25
Q

What is the definition of the partial sum, and how do we use this to find if the infinite series converges or diverges (2,2,1)

A

-Let an be a sequence
-The partial sum of the first K terms = ∑n=1k an

-The infinite series ∑k=1 ak converges is the partial sum (sk) converges to a finite limit as k -> ∞.
-We call this limit ‘S’ the sum of the series and write ∑k=1 ak = S

-If no such limit exists, we say the series diverges

26
Q

What is the sum rule for the convergence of 2 series (1)

A

-If the series’ ∑k=1 ak and ∑k=1 bk both converge, do does ∑k=1 ak + bk

27
Q

What are some tests for convergence (5)

A

-Sum rule
-Null sequence test
-Comparison test
-Ratio test
-Alternating series test

28
Q

What is the null sequence test for the convergence of series (1,2)

A

-The series ∑k=1 ak only converges if ak -> 0 as k -> 2

-This is a necessary but not sufficient condition for the convergence of a series
-For example, 1/k -> 0 as k -> ∞, but ∑k=1 1/k still doesn’t converge

29
Q

What is the comparison test for the convergence for series (1,3,1)

A

-Let ∑k=1 ak and ∑k=1 bk be series

If:
-ak, bk ≥ 0 for all k ∈ ℕ
-ak ≤ Mbk for all k ∈ ℕ and M > 0
-∑k=1 bk converges

-Then, ∑k=1 ak converges

30
Q

What is the comparison test for the divergence for series (1,3,1)

A

-Let ∑k=1 ak and ∑k=1 bk be series

If:
-ak, bk ≥ 0 for all k ∈ ℕ
-ak ≥ Mbk for all k ∈ ℕ and M > 0
-∑k=1 bk diverges

-Then, ∑k=1 ak diverges

31
Q

How can we prove that ∑k=1 1/k doesn’t converge (4)

A

-Let S = ∑k=1 1/k = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + …
-(1) = 1 = b1, (1/2) = 1/2 = b2, (1/3 + 1/4) > 1/2 = b3, (1/5 + 1/6 + 1/7 + 1/8) > 1/2 = b4
-This series is bounded below by the series ∑k=1 1/2
-Since ∑k=1 1/2 diverges by the null sequences test, therefore so does ∑k=1 1/k

32
Q

What is the ratio test for the convergence of series (2,2)

A

-Take a series ∑k=1 1
-Say (|ak+1/ak|) ->L
-If L < 1, ∑k=1 ak converges
-If L > 1, ∑k=1 ak diverges

33
Q

What is the alternating series test for the convergence of series (3,1)

A

For sequence ak, If:
-ak > 0 for all k ∈ ℕ
-ak is decreasing
-ak -> 0

-Then ∑k=1
(-1)k ak

34
Q

How can we prove series ∑k=1 (-1)k/k converges (1,3,1)

A

-Consider ∑k=1 1/k, which doesn’t converge

-However, 1/k > 0 for all k ∈ ℕ
-1/k -> 0 as k -> infinity
-This is thus a decreasing dequence

-Then, by the alternating series test, series ∑k=1 (-1)k/k converges