Overview (MA2730,2812,2815) lecture 15
Lecture slides for MA2730 Analysis I
Simon Shawpeople.brunel.ac.uk/~icsrsss
College of Engineering, Design and Physical Sciencesbicom & Materials and Manufacturing Research InstituteBrunel University
October 26, 2015
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Contents of the teaching and assessment blocks
MA2730: Analysis I
Analysis — taming infinity
Maclaurin and Taylor series.
Sequences.
Improper Integrals.
Series.
Convergence.
LATEX2ε assignment in December.
Question(s) in January class test.
Question(s) in end of year exam.
Web Page:http://people.brunel.ac.uk/~icsrsss/teaching/ma2730
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
MA2730: topics for Lecture 15
Lecture 15
Further tests for convergence
d’Alembert’s ratio test
Cauchy’s root test
Examples and Exercises
Reference: The Handbook, Chapter 4, Section 4.4.Homework: Finish all of Sheets 3a, 3b.Seminar: Proof of the ratio test. Q3 (parts) and Q4 on Sheet 3a.
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Where we are, and where we are going. . .
We understand a series as an infinite sum of a sequence.
We understand that partial sums of series generate sequences.
We understand that convergence is the central issue forseries. . .
. . . and that the value of a series is a secondary issue.
We have developed five tests for convergence.
Today we are going to develop two more:
d’Alembert’s ratio testCauchy’s root test
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
The geometric series — Reference: Stewart, Chapter 12.2.
We have seen this already, but here it is again.
n−1∑
k=0
ark =a(1− rn)
1− rif r 6= 1.
See Lecture 13, and Definition 4.5 in Subsection 4.1.3 of TheHandbook. Furthermore, we have also seen. . .
Theorem 4.6 in The Handbook
Let a and r be real numbers with |r| < 1. Then∞∑
k=0
ark =a
1− r.
Competence in recognising and using geometric series is essentialfor surviving undergraduate mathematics. Work at it and whenyou think you’ve got it, work at it again.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Study Habits and Time Management
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Don’t be a victim of your ownprocrastination.
Take control. Eat the frog.
Failure is easy. Challenge yourself.
Carpe Diem — sieze the day!
There are no refunds. Today counts.
Be ‘IN the room’
If you’re in the room, be IN the roomNigel Risner
Listen. Ask. Engage. Don’t chatter.
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
The ratio and root tests
Reference: Stewart, Chapter 12.6.
We’re ready now for our sixth and seventh convergence tests.Their proofs require the result for the geometric series.This material is drawn from Subsection 4.4.1 of The Handbook:The new tests are:
Theorem 4.22 — d’Alembert’s ratio test.
Theorem 4.23 — Cauchy’s root test.
The Handbook also gives Lemma 4.24 as the general form of bothof the above.
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Theorem 4.22 — d’Alembert’s ratio test
Suppose for the series∑∞
n=1 an we have
limn→∞
∣∣∣∣an+1
an
∣∣∣∣ = ρ
for some ρ > 0 (possibly infinite). Then:
if ρ < 1 the series converges.
if ρ > 1 or ρ = ∞ the series diverges.
If ρ = 1 this test provides no information.
We’ll defer the proof until we have seen the next test and someexamples.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Theorem 4.23 — Cauchy’s root test
Recall that |an|1/n = n√
|an|.
Suppose for the series∑∞
n=1 an we have
limn→∞
|an|1/n = ρ
for some ρ > 0 (possibly infinite). Then:
if ρ < 1 the series converges.
if ρ > 1 or ρ = ∞ the series diverges.
If ρ = 1 this test provides no information.
We will defer the proof for a moment and look at some examples.
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Examples
Use both the ratio and root tests on the following series.
1
∞∑
n=1
(n+ 3
2n+ 5
)n
.
2
∞∑
n=1
1
npfor p > 0.
3
∞∑
n=1
1
n!.
4
∞∑
n=1
xn
n!for x ∈ R.
Boardwork.
What do you notice about the last two?
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Before the proof, a very important and examinable result.
First, determine
limx→8
1
x− 8= ∞
BUT, what can you say about this?
limx→4
1
x− 4=
Back to business...
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Proof - the set up
For the ratio test we know that,
limn→∞
∣∣∣∣an+1
an
∣∣∣∣ = ρ
while for the root test, that
limn→∞
|an|1/n = ρ
where ρ > 0, and possibly infinite.We can write both of these in the form
limn→∞
bn = ρ
for bn =
∣∣∣∣an+1
an
∣∣∣∣ or bn = |an|1/n.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Proof - the key ingredient
So, for each theorem we know that
limn→∞
bn = ρ
and we have to see what we can conclude about∑
n an.First, let’s assume that ρ < ∞. Then, if bn → ρ as n → ∞ it mustbe the case that
|bn − ρ| can be made very small for large enough n
In mathematics we usually use the Greek ǫ, (‘epsilon’) to denote asmall positive number. So, we conclude,
|bn − ρ| < ǫ for all n > N
where N is a number that depends on how small ǫ is.Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Proof - the key insight
So, we are given thatlimn→∞
bn = ρ,
for bn =
∣∣∣∣an+1
an
∣∣∣∣ or bn = |an|1/n.
And when 0 6 ρ < ∞ we have concluded that
|bn − ρ| < ǫ for all n > N
where we can make ǫ as small as we please by choosing N largeenough.
This is DEEP — we’ll come back to it in the next lecture.
For now though, let’s just see what we get for this insight.Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Proof - the key result
For 0 6 ρ < ∞ we found that |bn − ρ| < ǫ for all n > N .We can make ǫ as small as we please by choosing N large enough.
Recall from level 1 that |bn − ρ| < ǫ means that −ǫ < bn − ρ < ǫ.
Or, by rearranging: ρ− ǫ < bn < ρ+ ǫ.
If ρ < 1 we can choose N so that ǫ < 12(1− ρ). This gives
bn < ρ+ ǫ < ρ+1
2(1− ρ) =
1
2(1 + ρ) < 1.
Hence: if ρ < 1 then bn < r where r = 12(1+ ρ) < 1 for all n > N .
Now we can write our proofs. First for the ratio test but. . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
The geometric series — Reference: Stewart, Chapter 12.2.
Let’s not forget this. . .
Theorem 4.6 in The Handbook
Let a and r be real numbers with |r| < 1. Then∞∑
k=0
ark =a
1− r.
Again:
Competence in recognising and using geometric series is essentialfor surviving undergraduate mathematics. Work at it and whenyou think you’ve got it, work at it again.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Revision from Lecture 15
Our first test. Lemma 4.14: absolute value convergence
If∑∞
k=1 |ak| converges, then∑∞
k=1 ak converges.
Comparison Test (third test): Theorem 4.17, Subsection 4.3.1
Let {ak} and {bk} be real non-negative sequences that satisfy
0 6 ak 6 bk for all k > N
for some N > 1. Then:
1 If∑∞
k=1 bk converges, so does∑∞
k=1 ak.
2 If∑∞
k=1 ak diverges, so does∑∞
k=1 bk.
Now to our proof. . .
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Proof of the ratio test: part 1
For the ratio test we had bn =∣∣∣an+1
an
∣∣∣ and we start with,
limn→∞
bn = limn→∞
∣∣∣∣an+1
an
∣∣∣∣ = ρ.
We have just deduced that if ρ < 1 then bn =∣∣∣an+1
an
∣∣∣ < r where
r = 12(1 + ρ) < 1 for all n > N .
Therefore, |an+1| < |an|r for all n > N . Iterating this:
n = N + 1 |aN+2| < |aN+1|r,n = N + 2 |aN+3| < |aN+2|r < |aN+1|r2,n = N + 3 |aN+4| < |aN+3|r < |aN+1|r3,n = N + 4 |aN+5| < |aN+4|r < |aN+1|r4 . . .
. . . and so on. Hence: |aN+k+1| < |aN+1|rk for all k > 1.Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Proof of the ratio test: part 2
If ρ < 1 we have |aN+k+1| < |aN+1|rk for r < 1 and all k > 1.Therefore, setting S =
∑Nn=1 |an| (a real number),
∞∑
n=1
an =
N∑
n=1
an+
∞∑
n=N+1
an 6N∑
n=1
|an|+∞∑
n=N+1
|an| 6 S+
∞∑
n=N+1
|an|.
But, using the geometric series,
∞∑
n=N+1
|an| =∞∑
k=0
|aN+k+1| 6∞∑
k=0
|aN+1|rk =|aN+1|1− r
.
Therefore∑ |an| converges and so, by Lemma 4.14, so does
∑an.
Essentially, this is the comparison test, Theorem 4.17.
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Revision from Lecture 14
We have just proven that if ρ < 1 the series∑
an converges.The next task is to prove that it diverges if ρ > 1.First, though, some revision from Lecture 14. . .
Theorem 4.7, Divergence Criterion
If∞∑
k=1
ak is convergent, then limk→∞
ak = 0.
Now back to the proof.
Can you see how theorems provide us with a tool box?
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Proof of the ratio test: part 3
For the ratio test we had bn =∣∣∣an+1
an
∣∣∣ and we start with,
limn→∞
bn = limn→∞
∣∣∣∣an+1
an
∣∣∣∣ = ρ.
Earlier we deduced that ρ− ǫ < bn for n > N .
So, if ρ > 1 we can choose N so that ǫ < 12(ρ− 1).
This gives bn > 1, and hence∣∣∣an+1
an
∣∣∣ > 1, for all n > N .
Therefore |an+1| > |an| for all n > N and so an 6→ 0 as n → ∞.
Therefore, by Theorem 4.7 (Lecture 14), the series∑
n an diverges.
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
That concludes the proof! Here, again, is what we have proven
Theorem 4.22 — d’Alembert’s ratio test
Suppose for the series∑∞
n=1 an we have
limn→∞
∣∣∣∣an+1
an
∣∣∣∣ = ρ
for some ρ > 0 (possibly infinite). Then:
if ρ < 1 the series converges.
if ρ > 1 or ρ = ∞ the series diverges.
If ρ = 1 this test provides no information.
Note the exclusion of the case ρ = 1 from our proof.
Now, we need to prove the root test. . .
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
This is what we had for Cauchy’s root test.
Theorem 4.23 — Cauchy’s root test
Recall that |an|1/n = n√
|an|.
Suppose for the series∑∞
n=1 an we have
limn→∞
|an|1/n = ρ
for some ρ > 0 (possibly infinite). Then:
if ρ < 1 the series converges.
if ρ > 1 or ρ = ∞ the series diverges.
If ρ = 1 this test provides no information.
Let’s look at the main steps in the proof — you’ll fill in the gapsfor homework.
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Proof of the root test
Put bn = |an|1/n then we have limn→∞
bn = |an|1/n = ρ.
Part 1: The case ρ < 1.
find r < 1 such that 0 6 bn < r and conclude that |an| 6 rn.
Conclude convergence by a geometric series comparison.
Part 2: The case ρ > 1.
find r > 1 such that 1 < r 6 bn and conclude that |an| > 1.
Use a previously established result to conclude divergence.
Homework
Attempt this for homework. We’ll look at it in next week’s seminar.
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
Lecture 15
Summary
We can:
Apply the ratio and root tests to series
Prove the ratio test theorem
Attempt a proof of the root test theorem
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MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 15
End of Lecture
Computational andαpplie∂ Mathematics
If you’re in the room, be IN the roomNigel Risner
Reference: The Handbook, Chapter 4, Section 4.4.Homework: Finish all of Sheets 3a, 3b.Seminar: Proof of the ratio test. Q3 (parts) and Q4 on Sheet 3a.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16