Infinite Series

9 downloads 294 Views 142KB Size Report
Engineering Mathematics: Open Learning Unit Level 1. 16.2: Sequences and Series. 2. Page 3. That is, Sn is the sum of the first n terms of the infinite series.








16.2

Infinite Series Introduction

We extend the concept of a finite series, met in Block 1, to the situation in which the number of terms increase without bound. We define what is meant by an infinite series being convergent by considering the partial sums of the series. As prime examples of infinite series we examine the harmonic and the alternating harmonic series and show that the former is divergent and the latter is convergent. We consider various tests for the convergence of series, in particular we introduce the Ratio test which is a test applicable to series of positive terms. Finally we define the meaning of the terms absolute and conditional convergence.



① be able to use the

Prerequisites Before starting this Block you should . . . ✫

Learning Outcomes After completing this Block you should be able to . . .





summation notation

② be familiar with the properties of limits ③ be able to use inequalities

Learning Style To achieve what is expected of you . . .

☞ allocate sufficient study time ✓ use the alternating series test and the ratio test on infinite series ✓ understand the terms absolute and conditional convergence

☞ briefly revise the prerequisite material ☞ attempt every guided exercise and most of the other exercises



1. Introduction Many of the series considered in Block 1 were examples of finite series in that they all involved the summation of a finite number of terms. When the number of terms in the series increases without bound we refer to the sum as an infinite series. Of particular concern with infinite series is whether they are convergent or divergent. For example, the infinite series 1 + 1 + 1 + 1 + ... is clearly divergent because the sum of the first n terms increases without bound as more and more terms are taken. It is less clear as to whether the harmonic and alternating harmonic series: 1+

1 1 1 + + + ... 2 3 4

1−

1 1 1 + − + ... 2 3 4

converge or diverge. Indeed you may be surprised to find that the first is divergent and the second is convergent. What we shall do in this Block is to consider some simple convergence tests for infinite series. Although we all have an intuitive idea as to the meaning of convergence of an infinite series we must be more precise in our approach. We need a definition for convergence which we can apply rigorously. First, using an obvious extension of the notation we have used for a finite sum of terms we denote the infinite series: a1 + a2 + a3 + . . . + ap + . . .

by the expression

∞ 

ap

p=1

where ap is an expression for the pth term in the series. So, as examples:

1 + 2 + 3 + ... =

∞ 

p

since the pth term is ap ≡ p

p=1 2

2

2

1 + 2 + 3 + ... =

∞ 

p2

since the pth term is ap ≡ p2

p=1 ∞  (−1)p+1 1 1 1 1 − + − + ... = 2 3 4 p p=1

here ap ≡

(−1)p+1 p

Consider the infinite series: a 1 + a2 + . . . + a p + . . . =

∞ 

ap

p=1

What we do is to consider the sequence of partial sums, S1 , S2 , . . . , of this series where S 1 = a1 S 2 = a1 + a2 .. . S n = a1 + a2 + . . . + a n Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

2

That is, Sn is the sum of the first n terms of the infinite series. If the limit of the sequence S1 , S2 , . . . , Sn , . . . can be found; that is lim Sn = S

(say)

n→∞

then we define the sum of the infinite series to be S: S=

∞ 

ap

p=1

and we say “the series converges to S”. Another way of stating this is to say that ∞  p=1

ap = lim

n→∞

n 

ap

p=1

Definition Convergence of Infinite Series An infinite series

∞ 

ap is convergent if the sequence of partial sums

p=1

S 1 , S 2 , S 3 , . . . , Sk , . . .

in which Sk =

k 

ap

is convergent

p=1

Divergence condition for an infinite series An almost obvious requirement that an infinite series should be convergent is that the individual terms in the series should get smaller and smaller. This leads to the following keypoint: Key Point The condition: ap → 0 as p increases (mathematically is a necessary condition for the convergence of the series

lim ap = 0)

p→∞

∞ 

ap

p=1

It is not possible for an infinite series to be convergent unless this condition holds.

3

Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

Now do this exercise Which of the following series cannot be convergent? (a) 12 + 23 + 34 + . . . (b) 1 + 12 + 13 + 14 + . . . (c) 1 − 12 + 13 − 14 + . . . In each case, use the condition from the previous Keypoint. Answer

Divergence of the Harmonic Series The harmonic series:

1 1 1 1 + + + + ... 2 3 4 5 has a general term an = n1 which clearly gets smaller and smaller as n → ∞. However, surprisingly, the series is divergent. Its divergence is demonstrated by showing that the harmonic series is greater than a series which is obviously divergent. We do this by grouping the terms of the harmonic series in a particular way:       1 1 1 1 1 1 1 1 1 1 1 1 + + + + + ... ≡ 1 + + + + + + + + ... 2 3 4 5 2 3 4 5 6 7 8 1+

Now

 1 1 1 1 1 + > + = 3 4 4 4 2   1 1 1 1 1 1 1 1 1 + + + > + + + = 5 6 7 8 8 8 8 8 2   1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 + + + + + + + > + + + + + + + = etc 9 10 11 12 13 14 15 16 16 16 16 16 16 16 16 16 2 

and so on. Hence the harmonic series satisfies:       1 1 1 1 1 1 1 1+ + + + + + + + ... 2 3 4 5 6 7 8       1 1 1 >1 + + + + ... 2 2 2 The right-hand side of this inequality is clearly divergent so the harmonic series is divergent

Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

4

Convergence of the Alternating Harmonic Series As with the harmonic series we shall group the terms of the alternating harmonic series, this time to display its convergence. The alternating harmonic series is: S =1−

1 1 1 1 + − + + ... 2 3 4 5

This series may be re-grouped in two distinct ways. 1st re-grouping 1 1 1 1 1 − + − + + ... = 1 − 2 3 4 5



1 1 − 2 3



 −

1 1 − 4 5



 −

1 1 − 6 7

 ...

each term in brackets is positive since 12 > 13 , 14 > 15 and so on. So we easily conclude that S < 1 since we are subtracting only positive numbers from 1. 2nd re-grouping 1 1 1 1 1 − + − + + ... = 2 3 4 5



1 1− 2



 +

1 1 − 3 4



 +

1 1 − 5 6

 + ...

Again, each term in brackets is positive since 1 > 12 , 13 > 14 , 15 > 16 and so on. So we can also argue that S > 12 since we are adding only positive numbers to the value of the first term, 12 . The conclusion that is forced upon us is that 1 a3 > . . . • the terms decrease to zero: ap → 0 as p increases

(mathematically

lim ap = 0)

p→∞

This is called the alternating series test. Now do this exercise Which of the following series are convergent (a)

∞ 

− 1) (−1) (2p + 1) p=1 p (2p

(b)

∞  (−1)p+1 p=1

p2 Answer

Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

6

3. The Ratio Test This test, which is one of the most useful and widely used convergence tests, applies only to series of positive terms.

Key Point The Ratio Test Let

∞ 

ap be a series of positive terms. Suppose, as p increases, the limit of

p=1

ap+1 equals ap

ap+1 = λ. Then, it is possible to show that: p→∞ ap

a number λ. That is lim

∞ 

• if λ > 1, then

ap diverges

p=1 ∞ 

• if λ < 1, then

ap converges

p=1 ∞ 

• if λ = 1, then

ap may be convergent or divergent.

p=1

That is, the test is inconclusive in this case.

Example Use the ratio test to examine the convergence of the series (a) 1 +

1 2!

+

1 3! 2

+

1 + 4! 3

...

(b) 1 + x + x + x + . . .

Solution (i) The general term in this series is

1 p!

i.e.

 1 1 1 1 + + + ... = 2! 3! p! p=1 ∞

and the ratio

7

ap =

1 p!



ap+1 =

1 (p + 1)!

ap+1 p! p(p − 1) . . . (3)(2)(1) 1 = = = ap (p + 1)! (p + 1)p(p − 1) . . . (3)(2)(1) (p + 1) Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

Solution (continued) ap+1 1 =0 = lim p→∞ ap p→∞ (p + 1)



lim

Since 0 < 1 the series is convergent. In fact, it will be shown in Block 4 that 1+

1 1 + + . . . = e − 1 ≈ 1.718 2! 3!

(ii) Here we must assume that x > 0 since we can only apply the ratio test to a series of positive terms. Now ∞  1 + x + x2 + x3 + . . . = xp−1 p=1

so that ap = xp−1 , and

ap+1 = xp

ap+1 xp = lim p−1 = lim x = x p→∞ ap p→∞ x p→∞ lim

Thus, using the ratio test we deduce that (if x is a positive number) this series will only converge if x < 1. We will see in Block 4 that 1 + x + x2 + x3 + . . .

=

1 1−x

provided 0 < x < 1.

Now do this exercise Use the ratio test to examine the convergence of the series: 27 8 1 + + ... + 2 ln 3 (ln 3) (ln 3)3 Answer Note that in all of these examples and guided exercises we have decided upon the convergence or divergence of various series; we have not been able to use the tests to discover what actual number the convergent series converges to.

4. Absolute and Conditional Convergence The ratio test applies to series of positive terms. Indeed this is true of many related tests for convergence. However, as we have seen, not all series are series of positive terms. To apply the ratio test such series must first be converted into series of positive terms. This is easily done. ∞ ∞   Consider two series ap and |ap |. The latter series, obviously directly related to the first, p=1

p=1

is a series of positive terms. Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

8

Using imprecise language, it is harder for the second series to converge than it is for the first, since, in the first, some of the terms may be negative and cancel out part of the contribution from the positive terms. No such cancellations can take place in the second series since they are ∞ ∞   all positive terms. Thus it is plausible that if |ap | converges so does ap . This leads to p=1

p=1

the following definition. Definition Conditional Convergence A convergent series

∞ 

ap for which its related series

p=1

∞ 

|ap | is divergent is said to be

p=1

conditionally convergent Absolute Convergence A convergent series

∞ 

ap is said to be absolutely convergent if

∞ 

p=1

|ap | is convergent.

p=1

For example, the alternating harmonic series: ∞  (−1)p+1 p=1

p

=1−

1 1 1 + − + ... 2 3 4

is conditionally convergent since the series of positive terms  ∞  ∞   (−1)p+1   1 1 1 ≡  = 1 + + + ...   p p 2 3 p=1 p=1 is divergent. Try each part of this exercise 1 1 1 Show that the series − + − + . . . is absolutely convergent. 2! 4! 6! Part (a) First, find the general term of the series Answer Part (b) Now obtain the related series of positive terms Answer Part (c) Now use the ratio test to examine the convergence of this series Answer  (p + 1)th term ? Part (d) What is lim p→∞ pth term 

Answer 9

Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

More exercises for you to try 1. Which of the following alternating series are convergent? (a)

∞  (−1)p ln(3)

(b)

p

p=1

∞  (−1)p+1 p=1

p2

+1

(c)

∞  p sin(2p + 1) π 2

p=1

(p + 100)

2. Use the ratio test to examine the convergence of the series: (a)

∞  p=1

(d)

∞  p=1

e4 (2p + 1)p+1 1 (0.3)p

(b)

∞  p3 p=1

(e)

p!

(c)

∞  1 √ p p=1

∞  (−1)p+1 p=1

3p

3. For what values of x are the following series absolutely convergent? (a)

∞  (−1)p xp p=1

p

(b)

∞  (−1)p xp p=1

p! Answer

Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

10

5. Computer Exercise or Activity

For this exercise it will be necessary for you to access the computer package DERIVE.

DERIVE will obtain the sum of many infinite series. For example to obtain the value of:  1 1 1 1 + 2 + 2 + ... = 2 3 p2 p=1 ∞

we would key Author: Expression 1/(p ∧ 2). DERIVE responds with 1 . p2 Now hit Calculus:Sum choosing p as the Variable and ∞ as the Upper limit. Hitting the OK button obtains the repsonse ∞  1 p2 p=1

Now choosing Simplify:Basic gets the response π2 6 We have considered, in this Block, the series 1 + x + x2 + . . . =

∞ 

xp

which, we found, was

p=1

only convergent if |x| < 1 (i.e. if −1 < x < 1). DERIVE will obtain the sum

∞ 

xp as long as

p=1

we constrain x to lie in the domain −1 < x < 1. This can be done, in DERIVE, by opening the Declare:Variable Domain menu. Type in Variable x and hit the Declare button. Then, in the screen that is presented, choose Real and Open interval. As Lower bound choose −1, and choose +1 for the Upper bound. Hit the OK button and DERIVE responds with x :∈ Real (−1, 1) which indicates that x is to be a real variable in the domain −1 < x < 1. Now construct

∞ 

xp

p=1

as you would normally in DERIVE. You will obtain the response: 1 1−x 11

Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

as expected. For some series, DERIVE will not be able to obtain its sum. However, DERIVE may still be used to check on the convergence or otherwise of the series. DERIVE can be used to apply the ratio test directly. To use this we need to use the Library MISC.MTH. To access this library of commands hit File:Open and double click on Misc icon. Then, for example, to apply the Ratio test to the series 1−

1 1 1 + + + ... 2 3 4

=

∞  (−1)p p=1

p

We would key in Ratio test ((−1) ∧ n/n , n) DERIVE responds:   (−1)n+1 , n RATIO TEST n Then hit the Simplify:Basic. DERIVE responds 1 as we expect. That is, in this case, the ratio test is inconclusive.

Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

12

End of Block 16.2

13

Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

p (a) ap = p+1 limp→∞ Hence series is divergent.

(b) ap =

1 p

p p+1

=1

lim ap = 0

p→∞

so this series may be convergent. Whether it is or not requires further testing. (c) ap =

(−1)p+1 p

lim ap = 0 so again this series may be convergent.

p→∞

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

14

(a) First, write out the series:

− 13 + 35 − 57 + . . .

1 (1 − 2p ) − 1) p Now examine the series for convergence. (−1) = (−1) → (−1)p as p 1 (2p + 1) (1 + 29 ) increases. Since the individual terms of the series do not converge to zero this is therefore a divergent series. p (2p

1 1 1 + 2 − 2 + . . . is an alternating series of the form 2 2 3 4 a1 − a2 + a3 − a4 + . . . in which ap = p12 . The ap sequence is a decreasing sequence since 1 1 1 > 2 > 2 > ... 2 3 1 Also lim 2 = 0. Hence the series is convergent by the alternating series test. p→∞ p

(b) This series 1 −

Back to the theory

15

Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series

First, find the general term of the series.

1 ln 3

+

8 (ln 3)2

+ ... =

∞  p=1

Now find ap+1

ap+1 =

p3 (ln 3)p

so ap =

p3 (ln 3)p

(p+1)3 (ln 3)p+1

ap+1 p→∞ ap

Finally, obtain lim

 3  3 1 Now p+1 = 1 + → 1 as p p p 1 ap+1 increases ∴ lim = p1 then the given series is divergent by comparison with the harmonic series. (d) λ = 10/3 so divergent, (e) Not a series of positive terms so the ratio test cannot be applied. ∞  |x|p 3. (a) The related series of positive terms is . For this series, using the ratio test we p p=1

find λ = |x| so the original series is absolutely convergent if |x| < 1. ∞  |x|p (b) The related series of positive terms is . For this series, using the ratio test we find p! p=1 λ = 0 (irrespective of the value of x) so the original series is absolutely convergent for all values of x. Back to the theory

21

Engineering Mathematics: Open Learning Unit Level 1 16.2: Sequences and Series