Schinazi R.B. From Calculus to Analysis

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60 2 Sequences and Series

converges to 0. By the root test, this series converges absolutely. We now ﬁnd a

numerical approximation. We have

=1 +

8923

6912

The error is



∞



n=1

−S



∞



n=5

For n ≥5, we have

≥5

and so

≤

Hence, by taking only the ﬁrst four terms the error is less than

∞



n=5

1 −1/5

2500

We now turn to alternating series.

Alternating Series Theorem

Assume that a

≥ 0 for all n ≥ 1, a

is a decreasing sequence, and that a

converges to 0. Then the series

∞



n=1

(−1)

converges.

Let S



k=1

(−1)

and E

. We have that

n+1

−E

2n+2

−S

=(−1)

2n+2

+(−1)

2n+1

2n+2

−a

2n+1

≤0,

where the inequality comes from the assumption that a

is a decreasing sequence.

Since E

n+1

−E

≤0, E

is also decreasing. We now show that it is bounded below.

Note that

=−a

≥−a

Similarly,

=−a

+(a

−a

) +a

≥−a

2.4 Absolute Convergence 61

since a

−a

≥0 and a

≥0. More generally, taking n ≥3, we have that

=−a

+(a

−a

) +(a

−a

) +···+(a

2n−2

−a

2n−1

) +a

≥−a

That is, the sequence E

is decreasing and bounded below by −a

; therefore, it

converges to some limit .LetF

2n−1

for n ≥1. Then

−F

≥0.

Since a

converges to 0, so does a

(why?), and by the squeezing principle, E

−F

converges to 0 as well. Therefore, F

converges to . At this point we have that S

and S

2n+1

converge to the same limit . We now show that this implies that the

whole sequence S

converges to .Takeany>0. There are N

and N

such that

if n ≥N

, then

−|<,

and if n ≥N

, then

2n−1

−|<.

Let N =max(N

) and take n ≥2N . Either n is even and n = 2k with k ≥N ≥

,so

−|<,

or n is odd and n =2k −1 with k ≥N ≥N

,so

2k−1

−|<.

We have shown that for any >0, there is 2N such that if n ≥2N, then

−|<.

That is, the alternating series

∞



n=1

(−1)

converges.

Numerical approximations are easy to compute for alternating series.

Numerical approximations

Assume that a

≥ 0 for all n ≥ 1, a

is a decreasing sequence, and that a

converges to 0. Then, for any natural k,wehave

2k−1



n=1

(−1)

≤

∞



n=1

(−1)

≤



n=1

(−1)

That is, if the alternating series theorem applies, then a partial sum gives an

approximation whose error is less than the ﬁrst omitted term.

62 2 Sequences and Series

In order to prove the double inequality above, we use the notation and the results

above. In particular, we have shown that E

= S

is decreasing and convergent.

Therefore, the limit  of E

(which is also the inﬁnite series) is less than E

for every

n ≥1. This proves the upper bound. For the lower bound, consider F

2n−1

.We

have

n+1

−F

=−a

2n+1

≥0

since a

is decreasing. That is, F

is increasing. We also know that F

converges

to the same . Therefore,  is larger than F

for all n ≥1. This provides the lower

bound above, and we are done.

Example 2.27 Consider



∞

n=1

(−1)

with a

= 1/n. Note that a

> 0 for all n,

is decreasing, and that it converges to 0. Therefore,



∞

n=1

(−1)

converges. More-

over, we have

=−1,S

=−1 +1/2 =−1/2,S

=−1/2 −1/3 =−5/6,

=−5/6 +1/4 =−7/12.

In particular,

−5/6 ≤

∞



n=1

(−1)

≤−7/12.

Exercises

1. Show that for every real x,wehave

0 ≤x +|x|≤2|x|.

2. Assume that the sequences a

and a

converge. Show that the sequence

converges as well.

3. Assume that a

=0 for all n.

(a) Show that if |a

n+1

|> |a

| for every n ≥N , then |a

|> |a

| for all n>N.

(b) Show that if the conclusion of (a) holds, then the sequence a

does not

converge to 0.

4. Does the series



n=1

(−1)

converge?

5. (a) Does the series



n=1

(2n)!

converge?

(b) Use (a) to ﬁnd the limit of the sequence

(2n)!

6. Does the series



n=1

converge?

7. Let a

=1/3

if n is not a prime and a

if n is a prime. Does the series



n=1

converge?

8. Under the assumptions of the alternating series theorem, show that

∞



n=1

(−1)

n−1

and

∞



n=1

(−1)

n+1

both converge.

2.4 Absolute Convergence 63

9. Find a numerical approximation for the series



∞

n=0

(−1)

10. Assume that |a

1/n

converges to some .

(a) Show that if  = 1, then the root test is not conclusive (you may use that

1/n

converges to 1).

(b) Show that if >1, then there is r>1 and a natural N such that if n ≥N,

then

|≥r

11. (a) Assume that a

, a

3n+1

, a

3n+2

all converge to the same limit . Show that

the sequence a

converges.

(b) Generalize the result in (a).

12. Assume that a

≥0 for every n and that the series



∞

n=1

converges.

(a) Show that



∞

n=1

(−1)

converges as well.

(b) Does (a) hold if the sequence a

is not assumed to be positive?

13. Assume that |

n+1

| converges to some limit >1.

(a) Show that there exist a real r>1 and a natural N such that for n>N,we

have that

|>r

n−N

(b) Show that |a

| goes to inﬁnity as n goes to inﬁnity. That is, show that for

every A, there is M such that if n ≥ M, then |a

|>A.

14. How many terms do we need to take to have a numerical approximation of



∞

n=1

with an error less than 10

−10

? Prove your claim.

15. (a) Find an upper bound for

∞



n=k+1

as a function of k.

(b) Find a bound on the error when



∞

n=1

is approximated by



n=1

16. In the exercises of Sect. 2.3 it was shown that a

≥0 for every n and that if the

series



∞

n=1

converges, then



∞

n=1

converges as well. Does this hold if

the sequence is not assumed to be positive?

17. (a) Assume that the series



∞

k=1

−a

k−1

|converges. Show that the sequence

converges.

(b) Suppose that for every k ≥1, we have

−a

k−1

|≤

Show that the sequence a

converges.

Chapter 3

Power Series and Special Functions

3.1 Power Series

A power series is a function deﬁned as a series.

Power series

A power series is a function f deﬁned by

f(x)=

∞



n=0

where c

is a sequence of real numbers.

Example 3.1 The simplest example of power series is given by the geometric series.

Recall from Example 2.10 in Sect. 2.2 that

∞



n=0

1 −x

for all |x|< 1

and that the series diverges for |x|≥1. Therefore, the power series

f(x)=

∞



n=0

1 −x

is deﬁned for x in (−1, 1). Note that the power series above is obtained when c

for all n ≥0.

The ﬁrst task when studying a power series is to decide where it is deﬁned. The

following result shows that the domain of a power series is always an interval.

R.B. Schinazi, From Calculus to Analysis,

DOI 10.1007/978-0-8176-8289-7_3, © Springer Science+Business Media, LLC 2012

66 3 Power Series and Special Functions

Radius of convergence

A power series



∞

n=0

has a radius of convergence R. The power series

converges if |x| <R, diverges if |x| >R, and can go either way if |x|=R.

We may have R =+∞, in which case the power series converges for all x.

Moreover, R is the least upper bound (if it exists!) of the set

I =



r ≥0 : the sequence c

is bounded



In particular, if |x|>R, then the sequence c

is not bounded.

To prove the result above, we ﬁrst need a lemma due to Abel.

Abel’s Lemma Assume that the sequence c

is bounded for a real b =0. Then

the power series



∞

n=0

converges absolutely for any x such that |x|< |b|.

We now prove Abel’s lemma. There is a real A such that for all n ≥1,



<A.

Take now x such that |x|< |b|.Forn ≥1, we have



The series



∞

n=0

converges since it is a geometric series with r =|

|< 1. By

the comparison test the (positive terms), series



∞

n=0

| converges as well. This

proves Abel’s lemma.

We now turn to the proof of the existence of the radius of convergence R.Let

I =



r ≥0 : the sequence c

is bounded



Note that the sequence c

is bounded by |c

| when r =0. Thus, 0 belongs to I ,

and I is not empty. There are two cases:

1. If I is bounded above, then, by the fundamental property of the reals, it has a

least upper bound R ≥ 0. If R = 0, then c

is not bounded for any x = 0.

By the divergence test, the power series converges for x = 0 only. Assume now

that R>0. For any x such that |x| <R, there must be at least one r in I such

that |x| <r<R (if all r in I are below |x|, then |x| is an upper bound of I

smaller than R, and that is not possible). By Abel’s lemma the series



∞

n=0

converges absolutely. This is true for any |x|<R. On the other hand, if |x|>R,

then |x| does not belong to I , and therefore c

|x|

is not bounded. So the power

series cannot converge at x for |x|>R.

2. If I is not bounded above, for any x, there is r such that r is in I and |x|<r.By

Abel’s lemma the power series converges absolutely for x. Thus, the power series

converges absolutely for all x.WesetR =+∞. This completes the description

of the domain of a power series.

3.2 Trigonometric Functions 67

The next example illustrates the fact that the ratio test is many times useful to ﬁnd

the radius of convergence of a series. It also shows that convergence at end points

can go either way.

Example 3.2 What is the domain of the power series



∞

n=1

(−1)

Let a

(−1)

. Then,

n+1

|x| converges to |x|/2. By the ratio test,

the power series converges if |x|/2 < 1 and diverges if |x|/2 > 1. Thus, the radius

of convergence for this series is 2. At this point we know that the domain of the

series includes (−2, 2), but it may or may not include one or both of the end points

of this interval.

We now examine the endpoints. Plugging x =2 into the series gives



∞

n=1

(−1)

which converges by the alternating series theorem. On the other hand, setting x =

−2 in the power series gives



∞

n=1

, which diverges (why?). Therefore, the domain

of this power series is (−2, 2].

Exercises

1. What is the radius of convergence of the series



∞

n=0

n!x

2. What is the domain of the series



∞

n=1

√

3. What is the domain of the series



∞

n=0

n+1

4. Assume that |c

n+1

|/|c

| converges to some limit .

(a) Show that if  = 0, then the radius of convergence of the series



∞

n=0

is inﬁnite.

(b) Show that if >0, then the radius of convergence of the series



∞

n=0

is 1/.

5. Assume that |c

n+1

|/|c

|goes to inﬁnity. That is, assume that for any A>0, there

is a natural N such that if n ≥N , then

n+1

|/|c

|>A.

Show that the power series f(x)=



∞

n=0

is deﬁned at x =0 only.

3.2 Trigonometric Functions

We start by deﬁning the functions sine and cosine using power series.

Sine and Cosine as power series

We deﬁne, for every real x,

sin x =

∞



n=0

(−1)

2n+1

(2n +1)!

cosx =

∞



n=0

(−1)

(2n)!

68 3 Power Series and Special Functions

For the deﬁnition above to make sense, the two power series must converge for

every x. We now check that. Let

=(−1)

2n+1

(2n +1)!

Then

n+1

=(−1)

n+1

2(n+1)+1

(2(n +1) +1)!

=(−1)

n+1

2n+3

(2n +3)!

and

n+1

(2n +1)!

(2n +3)!

|x|

2n+3

|x|

2n+1

(2n +2)(2n +3)

|x|

As n goes to inﬁnity,

n+1

converges to 0 < 1. By the ratio test the power series

converges absolutely for every x. Thus, the function sin is deﬁned everywhere. The

proof for cos is very similar and is left as an exercise for the reader.

First properties

We have cos0 =1 and sin 0 =0. The function cos is even, and the function

sin is odd.

The power series deﬁnition gives

sinx =x −

−

+···,

cosx =1 −

−

+···.

Letting x =0givessin0=0 and cos 0 =1. Since sin has only odd powers, we have

sin(−x) =−sinx. On the other hand, cos has only even powers, so cos(−x) =

cosx. This proves the ﬁrst properties.

If we differentiate term by term the power series deﬁning sinx, we get

(sinx)



=1 −

−

+···=1 −

−

+···.

Thus, it seems that the derivative of sin is cos. This is, of course, true. However,

in order to prove this, one needs to be able to differentiate a power series term by

term. Since there are inﬁnitely many terms in a series, this is not at all trivial. It will

be shown in Chap. 7 that a power series with radius of convergence R is inﬁnitely

differentiable on (−R,R) and that one can differentiate it term by term. Moreover,

the differentiated power series all have the same radius of convergence as the initial

power series.

We now do the same type of informal computation for cos, and we get

(cosx)



=−

−

+···=−



x −

+···



3.2 Trigonometric Functions 69

Again, this seems to indicate that (cosx)



=−sin x, and this is true.

Note that

(sin x)





(sinx)







=(cosx)



=−sin x.

Likewise,

(cosx)



=−cos x.

That is, sin and cos are both solutions of the differential equation f



+f = 0.

The next lemma will help us show that this differential equation characterizes sin

and cos.

Lemma Suppose that f is a function twice differentiable on R, that



+f = 0,

and that f(0) =f



(0) =0. Then is f is identically 0.

Deﬁne h =f

+(f



)

. Since f is differentiable, f

=f ·f is also differentiable.

The function f



is differentiable, so (f



)

is differentiable as well. Applying the

chain rule from Calculus gives



=2ff



+2f





=2f



(f +f



) =0,

where we use the hypothesis f



+f =0. A differentiable function whose derivative

is 0 on an interval is a constant on that interval. Thus, h is a constant. Since h(0) =0,

the function h is identically 0. Hence, h =f

+(f



)

is identically 0. Assume now,

by contradiction, that there is a real a such that f(a)=0. Then

h(a) =f(a)





(a)



≥f(a)

> 0.

Therefore, h is not identically 0. We have a contradiction. For all reals a, f(a)=0.

This completes the proof of the lemma.

Sine and Cosine as solutions of a differential equation

The function sin is the only solution of the differential equation



+f =0 with f(0) =0 and f



(0) =1.

The function cos is the only solution of the differential equation



+f =0 with f(0) =1 and f



(0) =0.

We now prove the claim above. Assume that there are two functions f and g

such that



+f =0 with f(0) =0 and f



(0) =1,



+g =0 with g(0) =0 and g



(0) =1.