RAC sequences and series Flashcards
sequence (aₙ)
an ordered of real numbers
series
the sum of all the terms in a sequence
trichotomy
exactly one of the following statements is true:
a<b></b>
a>b =>
a + c > b + c
if a>b and c>0 =>
ac > bc
if a>b and c<0 =>
ac < bc
a>b>0 =>
1/a < 1/b
if ab>0 <=>
a,b>0 or a,b<0
using a sign table
- split the expression
- decide which values cause any of the expressions to be 0
- write the sign for each expression and each value including the ‘ general’ inbetween values
- depending on whether or not its a strict inequality write the intervals which match
properties of the modulus function
- |x|>= 0 and |x| =0 <=> x=0
- since |x|^2 = x^2, so |x| =√(x^2)
- |x|<=x<=|x|
- if Ɛ>0, |x| -Ɛ
|a-b|
the distance between a and b
triangle theorum
|a+b| <= |a| + |b| for all real numbers
Tending to infinity definition
A sequence (aₙ) tends to infinity if given any real number, A>0, there exists a point in the sequence, NЄℕ, such that an>A wherever n>N
Tending to minus infinity definition
A sequence (aₙ) tends to minus infinity if given any real number, A>0, there exists a point in the sequence, NЄℕ, such that anN
(aₙ) -> -∞ <=>
-(aₙ) -> ∞
Converging to a point definition
A sequence (aₙ) converges to a real number l if: given any small number, Ɛ>0, there exists a point in the sequence, NЄℕ, such that |(aₙ)-l|N
uniqueness of limits theorum
if a sequence converges then its limit, l, is unique
proving something tends to infinity
- Let A>0
- simplify the expression to observe that (expression for n)>A
- rearrange for n>(expression of A)
- provided n> expressions. choosing any natural number larger than expression we have an>A for all n>N
triangle theorem
|a+b| <= |a| + |b| for all real numbers
Suppose (an) converges to l>0. then there exists NЄℕ such that…
an> 0 whenever n>N
bounded above definition
there exists MЄℝ such that an<=M for all nЄℕ
bounded below definition
there exists MЄℝ such that an>=M for all nЄℕ
bounded definition
both bounded above and bounded below
if a sequence converges…
it is bounded. if a sequence is bounded it doesn’t mean it converges.
Algebra of limits theorem
suppose (an) and (bn) are sequences converging to l and M, then:
- an + bn -> l + M
- λan -> λl
- anbn -> lM
- an/bn -> l/M
Converging to a point definition
A sequence (aₙ) converges to a real number l if: given any small number, Ɛ>0, there exists a point in the sequence, NЄℕ, such that |(aₙ)-l|N
Tending to infinity definition
A sequence (aₙ) tends to infinity if given any real number, A>0, there exists a point in the sequence, NЄℕ, such that an>A wherever n>N
tending to minus infinity definition
A sequence (aₙ) tends to minus infinity if given any real number, A>0, there exists a point in the sequence, NЄℕ, such that anN
proving something converges to a limit
- let Ɛ>0
- sub in an and l into (aₙ)-l
- rearrange in terms of l
- Given N is any number at least as large as (expression in terms of Ɛ) we have (aₙ)-lN
proving a complicated expression converges to a point
- (observe that) get in terms of 1/n for each value (remove other values if neccessary)
- using algebra of limits see what each term tends to
- work out the overall point of convergence
what does 1/n converge to?
0
proving algebra of limits theorems
- use convergence definition for each sequence/limit for all n>N1
- use triangle inequality
- take maximum of expression
limiting limits of sequences proposition
an->l bn->m
if an<=bn then l
sandwhich theorem
an<=bn<=cn
if an-> and cn -> l then bn -> l
subsequence
selecting only certain terms from an original sequence
convergence of a sub-sequence theorem
suppose an->l and ank is a susequence then ank->l
showing a sequence doesn’t converge
if the sequence possesses subsequences converging to distinct limits then it does not converge
Bolzano wierstrass Theorem
Every bounded sequence of real numbers includes a convergent subsequence
Monotone increasing sequence
an+1>=an
Monotone strictly increasing sequence
an+1>an
Monotone decreasing sequence
an+1<=an
Monotone strictly decreasing sequence
an+1
Montone sequence
is either increasing, decreasing, strictly increasing or strictly decreasing
Monotone convergence theorem
- if an is increasing and bounded above then it converges
2. if an is decreasing and bounded below then it converges
eulers theorem
a sequence an given an=(1+1/n)^n converges
A series is an expression of the form
Σaₙ = a₁+a₂+a₃+a₄+…
method of differences
- split the expression into individual fractions
- subbing in values
- cancelling
- left with and expression in terms of N
converging series
a series Σaₙ converges to a real number s if its sequence of partial sums Sₙ = Σaₙ converges to S
diverges
if a series does not converge it diverges
geometric series
Σrⁿ= 1 + r + r²+…+rⁿ
sum of a geometric series
Sₙ = (1-r⁽ⁿ⁺¹⁾)/1-r if r!=0
infinite sum of a geometri series
Sₙ = 1/(1-r) if |r|<1
a geometric series converges iff
|r|<1
Σ1/nᵃ converges
if a>1
Algebra of limits for series theorem
suppose Σaₙ and Σbₙ converge and λ,μ are real then Σλaₙ + μbₙ converges and Σλaₙ + μbₙ = λΣaₙ + μΣbₙ
Sequence Convergence Tests
- Null sequence test
- The comparison test
- The ratio test
- The root test
- The Integral test
- Alternating Series Test
- Absolute Convergence Test
Null Sequence Test
if Σaₙ converges, then an->0
test for divergence only: if an->0 then Σaₙ diverges
The Comparison Test
Suppose an>=0 bn>=an for n then:
- Σbₙ converges <=> Σaₙ converges
- Σaₙ diverges <=> Σbₙ diverges
The Ratio Test
Suppose Σaₙ is a series of non-negative terms and: aₙ₊₁/aₙ -> r then: 1. if r<1 => Σaₙ converges 2. if r>1 Σaₙ diverges 3. if r=0 inconclusive
The root test
Suppose Σaₙ is a series of non-negative terms and aₙ¹/ⁿ->. Then:
- if r<1 => Σaₙ converges
- if r>1 => Σaₙ diverges
- if r=0 inconclusive
Integral Test
Suppose f:[1,∞)-> is continuous, decreasing and positive in its domain. Then the series Σfₙ converges if and only if the sequence (∫f(x)dx) converges
Alternating series test
Consider the series Σ(-1)ⁿ⁺¹aₙ where the sequence (aₙ) is decreasing and converging to zero. Then the series converges.
Absolute Convergence Test
If a series converges absolutely absolutely, then it converges
absolute convergence
Σaₙ converges absolutely if Σ|aₙ|
conditional convergence
Σaₙ converges conditionally if it converges, but not absolutely
rearrangement
we a sequence bₙ a rearrangement of a sequence an if there is a bijection σ:ℕ-> ℕ such that bₙ = aσ₍ₙ₎
Dirichlets theorem (rearrangement of series)
Let Σaₙ be absolutely convergent. if bₙ is any rearrangement of aₙ then:
- Σbₙ is absolutely convergent
- Σbₙ = Σaₙ
Conditional convergence theorem
suppose Σaₙ is conditionally convergent. Given γ there exists a rearrangement bₙ of aₙ such that Σbₙ=γ
Multiplying series theorem
if Σaₙ and Σbₘ are Absolutely convergent then Σaₙbₘ is well-defined (order doesn’t matter) and absolutely convergent and: (Σaₙ) * (Σbₘ) = Σaₙbₘ
form of a power series
Σaₙxⁿ
convergence of power series theorem
Given a power series Σaₙxⁿ, either it converges absolutely for all x∈ℝ, or there exists ℝ∈[0,∞) such that
