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MATH10242 SUMMARY
Sequences
The Triangle Inequality
For all x and y, we have
|x + y| ≤ |x| + |y|
Convergence
A sequence (an )n∈N converges to l if:
∀ > 0, ∃N ∈ N such that
|an − l| < holds for all n ∈ N with n ≥ N
So, even when is really small, you can still find N ∈ N such that all
the terms after the N th term of the sequence have a difference with the limit
l that is less than A sequence can converge to at most one limit
Cauchy Sequence
A sequence (an )n∈N is a Cauchy sequence if ∀ > 0, ∃N ∈ N such that
|ai − aj | < holds for all i, j ∈ N with i, j ≥ N
So terms of a Cauchy sequence are getting closer and closer
A sequence converges ⇔ it is a Cauchy sequence
Bounded
A sequence (an )n∈N is bounded if
∃M ∈ R+ , ∀n ∈ N, |an | ≤ M
So all the terms of the sequence are greater than −M and less than M
A sequence (an )n∈N is convergent ⇒ (an )n∈N is bounded.
Monotone Convergence Theorem
If a sequence (an )n∈N is both increasing and bounded, then (an )n∈N is
convergent
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The Sandwich Rule
If sequences (an )n∈N and (bn )n∈N both converge to l ∈ R and we have sequence
(cn )n∈N such that for large n,
(an )n∈N ≤ (cn )n∈N ≤ (bn )n∈N
then cn → l as n → ∞
The Algebra of Limits
Let sequences (an )n∈N and (bn )n∈N converge to real numbers a and b respectively,
then
(modulus)
limn→∞ |an | = |a|
(scalar multiplication) for any k ∈ R, limn→∞ kan = ka
(addition)
limn→∞ (an + bn ) = a + b
(multiplication)
limn→∞ (an · bn ) = a· b
(division)
if bn 6= 0 (for all n and b 6= 0, then
a
an
=
limn→∞
bn
b
(Reciprocal)
if bn 6= 0 (for all n and b 6= 0, then
1
1
limn→∞ =
bn
b
Divergence
A sequence (an )n∈N is divergent if it is not convergent, so there is no l ∈
R such that
an → l as n → ∞
an example of a divergent sequence is ((−1)n )n∈N
note that a sequence can be divergent even if it is bounded
Tends To Infinity
We say a sequence (an )n∈N tends to ∞ as n → ∞ if ∀K ∈ R> , ∃N ∈ N such
that
an > K ∀n ≥ N
The Reciprocal Rule
Let (an )n∈N be a sequence of non-zero real numbers, then:
(1) an → ∞ as n → ∞
⇒
(2) an > 0 for large n and
1
an
is null
⇒
1
→ 0 as n → ∞
an
an → ∞ as n → ∞
n∈N
A sequence is null if it converges to 0
If (an )n∈N and (bn )n∈N both tend to infinity, then
(1) an + bn → ∞ as n → ∞
(2) for c > 0, c· an → ∞ as n → ∞
(3) an · bn → ∞ as n → ∞
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Infinity Sandwich Rule
If (bn )n∈N → ∞ as n → ∞ and (an )n∈N is any sequence such that for all large
n,
an ≥ b n
then an → ∞ as n → ∞
Subsequences
For any subsequence (akn )n∈N of the sequence (an )n∈N :
(1) an → l as n → ∞
⇒ akn → l as n → ∞
(2) an → ∞ as n → ∞
⇒ akn → ∞ as n → ∞
(3) an → −∞ as n → ∞ ⇒ akn → −∞ as n → ∞
Bolzano-Weierstrass Theorem (1817)
Every bounded sequence (an ) has a convergent subsequence
L’Hˆ
opital’s Rule
Given the Conditions:
(1) f : [1, ∞) → R and g : [1, ∞) → R are two functions and can
be differentiated at least twice
(2) for some N > 0, g(x) > 0 for x > N
(3) (an )n∈N and (bn )n∈N are sequences such that (an ) = f (n)
and (bn ) = g(n) for n ∈ N
(4) n→∞
lim an = n→∞
lim bn = ∞
Then we have
lim
n→∞
f 0 (n)
an
= n→∞
lim 0
bn
g (n)
Series
Given the sequence of partial sums (sn )n∈N such that
s n = a1 + a2 + · · · + an =
n
X
ai
i=1
If sn → s as n → ∞, then
∞
P
an is convergent with sum s.
n=1
If there is no s ∈ R with this property, then
∞
P
n=1
∞
P
n=1
an is divergent.
an is a convergent series ⇒ (an )n∈N is a null sequence.
The Algebra of Infinite Sums
Let series
∞
P
n=1
an and
∞
P
n=1
bn be convergent with sums A and B respectively, then
(addition)
∞
P
(an + bn ) is also convergent with sum A + B
n=1
(scalar multiplication)
for any λ ∈ R,
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∞
P
n=1
λan is convergent with sum λA
Series With Non-negative Terms
Let sn =
n
P
i=1
ai such that an ≥ 0 for all n ∈ N, then
the series
∞
X
an is convergent ⇔ the sequence (sn )n∈N is bounded above
n=1
Each ai is non-negative, so sn =
n
P
ai is increasing and hence the sequence
i=1
(sn )n∈N is increasing.
This is similar to the Monotone Convergence Theorem.
The Comparison Test (improved)
Given series
∞
P
n=1
an and
N > 0), then
∞
P
n=1
∞
X
bn such that 0 ≤ an ≤ bn for all n > N (for some
bn converges ⇒
n=1
∞
X
an converges,
n=1
∞
X
an diverges ⇒
n=1
∞
X
bn diverges
n=1
The original comparison test requires condition such that 0 ≤ an ≤ bn for
all n ∈ N, but since infinite sums have the (very important!) property that
∞
X
an is convergent ⇔
n=1
X
an is convergent for some positive N ∈ N
n≥N
so the improved comparison test works. This property can be applied in
conjunction with other tests for series convergence.
The Ratio Test For Series
an+1
→ l as n → ∞
an
Let an > 0 for all N ∈ N and assume
(1) l < 1 ⇒
(2) l > 1 ⇒
∞
P
= an is convergent
n=1
∞
P
= an is divergent
n=1
However, no conclusion can be drawn if l = 1
The Integral Test
Let f : [1, ∞) → R be a function that is positive (f (x) ≥ 0 ∀x ≥ 0), decreasing
(f (x) ≥ f (y) ∀x ≥ y ∈ R) and continuous, then
the series
∞
X
f (n) converges ⇔
n=1
Z n
1
n∈N
f (x)dx
1
converges as n → ∞ ⇔
f (x)dx
the sequence
Z n
is bounded (monotone convergence theorem)
n∈N
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Series With Positive And Negative Terms
The Alternating Series Test
If (an )n∈N is a decreasing null sequence with positive terms, then
the series
∞
X
(−1)n+1 an is convergent.
n=1
I think for (an )n∈N to be null and positive implies that it is also
decreasing after the N th term for some positive integer N (that’s if
it’s not decreasing from the beginning).
Absolute Convergence
A series
If
∞
P
n=1
∞
P
n=1
an is absolutely convergent if
∞
P
n=1
|an | is convergent.
an is convergent but not absolutely convergent, then it is
conditionally convergent
∞
X
|an | is convergent ⇒
∞
X
an is convergent
n=1
n=1
THIS IS A SUMMARY OF MATH10242 UPTO WEEK 10,
ITS PURPOSE IS ONLY TO JOG YOUR MEMORY. PLEASE
READ THE PROPER NOTES WHEN REVISING.
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