Lecture 15 - Brunel University Londonicsrsss/teaching/ma2730/lec/print8lec...procrastination. Take...

Overview (MA2730,2812,2815) lecture 15

Lecture slides for MA2730 Analysis I

Simon Shawpeople.brunel.ac.uk/~icsrsss

[email protected]

College of Engineering, Design and Physical Sciencesbicom & Materials and Manufacturing Research InstituteBrunel University

October 26, 2015

Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel

MA2730, Analysis I, 2015-16


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




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.




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




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.




Lecture 15

Study Habits and Time Management




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.




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.




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.




Lecture 15

Theorem 4.23 — Cauchy’s root test

Recall that |an|1/n = n√

|an|.


n=1 an we have

limn→∞

|an|1/n = ρ





We will defer the proof for a moment and look at some examples.




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?




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...




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.




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



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



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. . .




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.




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. . .




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



Lecture 15


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.




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?




Lecture 15


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.




Lecture 15

That concludes the proof! Here, again, is what we have proven

Theorem 4.22 — d’Alembert’s ratio test


n=1 an we have

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = ρ





Note the exclusion of the case ρ = 1 from our proof.

Now, we need to prove the root test. . .




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|.


n=1 an we have

limn→∞

|an|1/n = ρ





Let’s look at the main steps in the proof — you’ll fill in the gapsfor homework.




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.




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




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.



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