Math 23a
Section 6
October 19th, 2015
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Week 6 Roadmap
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Key Concepts in Brief
• A sequence (sn ) is increasing if sn ≤ sn+1 , ∀n. This changes to strictly
increasing if sn < sn+1 , ∀n. Analogous definitions follow for decreasing
and strictly decreasing.
• These sequences, in general, are called monotone sequences
• VERY IMPORTANT THEOREM: All bounded monotone sequences
converge to the supremum or infimum, depending on if it’s increasing or
decreasing.
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• A sequence is Cauchy if
∀ > 0 ∃N s.t. ∀m, n > N, |sn − sm | < This is a very important definition!
• All convergent sequences are Cauchy. Cauchy sequences of real numbers are convergent. These are also two very important theorems. Proving
a sequence of real numbers to be Cauchy is perfectly valid to prove convergence, and it does not require knowing the limit beforehand.
• The lim sup is the limit of the supremum of the remaining elements in the
set as N goes to infinity. In mathematical notation, this is
lim sup sn = lim sup{sn : n > N }
N →∞
• The lim inf is the limit of the infimum of the remaining elements in the set
as N goes to infinity. In mathematical notation, this is
lim inf sn = lim inf{sn : n > N }
N →∞
• If the limit exists, then
lim inf sn = lim sup sn = lim sn
Likewise, you can assert that if lim inf sn and lim sup sn exist and are equal,
then the limit exists and is exactly the value of lim inf sn or lim sup sn
• A subsequence is created from a parent sequence by selecting an infinite
number of terms from that sequence in order. If the parent sequence converges to s, all subsequences converge to s.
• The Bolzano-Weierstrass Theorem will turn out to be incredibly useful,
and it states that any bounded sequence has a convergent subsequence.
(PROOF 6.1)
• The partial sum is formed from a sequence by adding together the terms
up to a certain point. This is notated as:
sn =
n
X
ak
k=m
Be very careful with this notation, we are talking about a sum of terms of
a sequence, yet it has been notated here with s. Usually the lower bound
of summation is 0 or 1.
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P
• The series,
an is said to converge if lim sn = S. It is said to diverge, if
the limit is ∞ or −∞.
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• A series is absolutely convergent if the series
|an | converges as well.
• The geometric series a + ar + ar2 + ... converges if |r| < 1, and converges
a
to 1−r
∞
X
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• A p-series,
, converges if p > 1, and diverges otherwise.
np
n=1
• The Cauchy Criterion basically states that if the sequence of partial sums
are Cauchy, then it converges. This can be recast as
∀ > 0, ∃N s.t. ∀n ≥ m > N, |
n
X
ak | < k=m
• The convergence tests are your go to methods to see if a series converges
or diverges.
P
– ComparisonP
Test: Consider the series
an . If bn <
P of positive terms
a
∀n,
and
a
converges,
then
b
converges.
If
b
>
a
n
n
n
n
n ∀n, and
P
P
an diverges, then
bn diverges
P
|<
– Ratio Test: Consider the series of non zero terms an . If lim sup | an+1
an
an+1
1, then the series converges. It diverges if lim inf | an | > 1. It is inconclusive otherwise.
P
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– Root Test: Consider the series
an . If lim sup |an | n < 1, then it
converges. If it is greater than 1 it diverges. If it is 1, it is inconclusive.
(PROOF 6.2)
– Alternating Series Test: If we have an alternating series and the limit
of the absolute value of the terms go to zero, then the alternating series
converges.
• To prove that a series diverges, just show that the terms do not converge
to zero. If they do converge to zero, then another test must be used.
P
1
n
• In a power series with the form ∞
n=0 an x , the radius of convergence is β ,
1
where β = lim sup |an | n
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Key Definitions
3.1
Cauchy Sequence
This definition is absolutely critical to remember! It is
∀ > 0 ∃N s.t. ∀n, m > N, |sn − sm | < 3.2
lim sup and lim inf
These two are also very important definitions to remember, because they are
involved in convergence tests and can be used to show that a sequence converges.
We’ll only discuss the lim sup, because lim inf is completely analogous. A key
thing to remember here also is that lim sup an ≤ sup an , and that lim inf an ≥
inf an . We can see this graphically.
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Using the Convergence Tests
Here are some guidelines with how to use these convergence tests that we have
at our disposal effectively.
• If there are factorials, almost always use the ratio test. This is because,
factorials do not lend themselves to having roots taken at all
• If there is anything else, use the root test. The root and ratio test will
actually cover almost all possible questions you may be asked.
• What to do if there are these extra additive or multiplicative factors that
seem to hinder our root or ratio test? Then, use the comparison test! The
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comparison test can simplify our problems, and usually involves reasoning
such as:
– “If I make the denominator smaller by getting rid of this stupid +50n2 ,
then surely my new sequence will get larger, and if that converges, then
my original sequence converges.”
– “If I make my numerator smaller by getting rid of this ridiculous cosine
and replacing it with +2, then surely my new sequence will get larger,
and if that converges, then my original sequence converges.”
– “If I make my denominator larger by replacing this n − 1 with n, then
surely my new sequence is smaller, and if this sequence diverges, then
my original sequence diverges!
4.1
Does
Examples
∞
X
1 + sin(n)
n=2
Does
5
log(n)
converge?
X cos2 (n) − 3√n
n2 + πn
converge?
Problems
1. Prove that if the limit of a sequence is s, then the limit of every subsequence
of that sequence is also s.
2. Ross Exercise 10.7 Let S be a bounded nonempty subset of R such that
sup S is not in S. Prove there is a sequence (sn ) of points in S such that
lim sn = sup S.
3. Ross Exercise 12.10 Prove (sn ) is bounded if and only if lim sup |sn | < +∞.
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1. A subsequence is an infinite selection of elements from the parent sequence,
sn . Let’s call this subsequence tk where tk = sn where k ≤ n. Because the
sequence converges to s, choose , and N , such that ∀n > N |sn − s| < .
Because tk = sn where k ≤ n, then the following also holds ∀n > N |tn −
s| < , because tn is included in the ∀n > N assertion, because tn is at
worst going to equal sn , but is most likely gonna be farther along in the
sequence, which will fall under the convergence assertion. Therefore, any
subsequence tn will converge to s.
2. By the completeness axiom, we can assert the existence of the sup S = M .
We know that ∀n ∈ N , M − n1 cannot be a upper bound, because M is the
least upper bound. Therefore ∃sn s.t. M − n1 < sn < M . By the Squeeze
Theorem, and taking the limit of each parts of this inequality to infinity, we
obtain lim sn = M . Alternatively, you can imagine, forming each element of
sn , from the given lower bound of M − n1 . By construction, sn is increasing,
and it is also bounded, so therefore it must converge to M .
3. Because this is a if and only if proof, otherwise known as a biconditional
proof, we must prove both directions. First, the easy way, let’s prove that if
sn is bounded, then the lim sup |sn | < +∞. Call the sup |sn | = M < +∞.
The lim sup |sn | ≤ sup |sn | < +∞. Now, let’s prove that if lim sup |sn | <
+∞, then the sequence sn is bounded. Using Ross’s notation, let vN =
sup{sn : n > N }, therefore lim sup sn = lim vN = A. Choose > 0, and by
convergence, ∃M s.t. ∀N > M |vN − A| < =⇒ vN < A + =⇒ ∀n >
N, sn < A+. Therefore, we have our bound M as max(s1 , s2 , ..., sn , A+).
This is a finite number of elements, and thus the maximum is finite, and
the sequence is bounded.
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