2 Sequences and Series of Functions

If ( f

) is a sequence of functions, what should we mean by lim f

? This question is of great relevance

to the history of calculus; Issac Newton’s work in the late 1600’s made great use of power series, which

are naturally constructed as limits of sequences of polynomials.

For instance, for each n ∈ N

, we might consider the polynomial function f

: R → R deﬁned by

(x) =

∑

k=0

= 1 + x + ··· + x

This is easy to work with, to differentiate and integrate using the power law. What, however, are we

to make of the following series?

f (x) :=

∞

∑

n=0

= 1 + x + x

+ ···

Does this make sense? What is its domain? Does it equal the limit of the sequence ( f

) in a meaning-

ful way? Is it continuous, differentiable or integrable? Can we compute its limit/derivative/integral

term-by-term in the obvious way; for instance, is it legitimate to write

′

(x) =

∞

∑

n=1

n−1

= 1 + 2x + 3x

+ ···?

To many in Newton’s time, these questions were of diminished importance when compared to the

burgeoning applications of calculus to the natural sciences. However, for the 18

and 19

century

mathematicians who followed, the widespread application of calculus only increased the imperative

to rigorously address these issues.

23 Power Series

First we recall some of the important deﬁnitions, examples and results regarding inﬁnite series.

Deﬁnition 2.1. Let (b

)

∞

n=m

be a sequence of real numbers. The (inﬁnite) series

∑

is the limit of

the sequence (s

) of partial sums,

∑

k=m

= b

+ b

m+1

+ ···+ b

∞

∑

n=m

= lim

n→∞

The series

∑

converges, diverges to inﬁnity or diverges by oscillation

if the sequence (s

) does so.

∑

is absolutely convergent if

∑

converges. A convergent series that is not absolutely convergent

is conditionally convergent.

Recall that every sequence (s

) has subsequences tending to each of

lim sup s

= lim

N→∞

sup{x

: n > N} and lim inf s

= lim

N→∞

inf{x

: n > N}

If (s

) converges, or diverges to ±∞, then lim s

= lim sup s

= lim inf s

. The remaining case, divergence by oscillation,

is when lim inf s

= lim sup s

: there exist (at least) two subsequences tending to different limits.

Examples 2.2. These examples form the standard reference dictionary for analysis of more complex

series. Make sure you are familiar with them!

1. (Geometric series) Let r be a constant, then s

∑

k=0

1−r

n+1

1−r

. It follows that

∞

∑

n=0











converges (absolutely) to

1−r

if −1 < r < 1

diverges to ∞ if r ≥ 1

diverges by oscillation if r ≤ −1

2. (Telescoping series) If b

n(n+1)

, then s

∑

k=1

= 1 −

n+1

=⇒

∞

∑

n=1

n(n+1)

= 1.

∞

∑

n=1

is (absolutely) convergent. In fact

∞

∑

n=1

, though checking this explicitly is tricky.

4. (Harmonic series)

∞

∑

n=1

is divergent to ∞.

5. (Alternating harmonic series)

∞

∑

n=1

(−1)

is conditionally convergent.

Theorem 2.3 (Root Test). Given a series

∑

, let β = lim sup

1/n

• If β < 1 then the series converges absolutely.

• If β > 1 then the series diverges.

We give sketch proofs, and/or refer you to a standard ‘test.’ Review these if you are unfamiliar.

1. s

−rs

= 1 + r + ··· + r

−(r + ··· + r

+ r

n+1

) = 1 − r

n+1

=⇒ s

1−r

n+1

1−r

2. b

−

n+1

=⇒ s



1 −





−



+ ··· +



−

n+1



= 1 −

n+1

3. Use the comparison or integral tests. Alternatively: For each n ≥ 2, we have

n(n−1)

. By part 2,

∑

k=1

< 1 +

∑

k=1

k(k −1)

≤ 2

Since (s

) is a monotone up sequence, bounded above by 2, we conclude that

∑

is convergent.

4. Use the integral test. Alternatively, observe that

n+1

−s

n+1

∑

k=2

−1

≥

n+1

=⇒ s

≥

−−−→

n→∞

∞

Since s

∑

k=1

deﬁnes an increasing sequence we conclude that s

→ ∞.

5. Use the alternating series test, or explicitly check that both the even and odd partial sums (s

) and (s

2n+1

) are

convergent (monotone and bounded) to the same limit.

Root Test: β < 1 =⇒ ∃ϵ > 0 such that

1/n

≤ 1 − ϵ (for large n) =⇒

∑

converges by comparison with the

convergent geometric series

∑

(1 − ϵ)

β > 1 =⇒ a subsequence of (

1/n

) converges to β > 1, whence b

↛ 0 =⇒

∑

diverges (n

-term test).

The root test is inconclusive if β = 1. Some simple inequalities

yield a simpler test.

Corollary 2.4 (Ratio Test). Given a series

∑

• If lim sup



n+1



< 1 then

∑

converges absolutely.

• If lim inf



n+1



> 1 then

∑

diverges.

We can now properly deﬁne and analyze our main objects of interest.

Deﬁnition 2.5. A power series centered at c ∈ R is a formal expression

∞

∑

n=m

(x −c)

where (a

)

∞

n=m

is a sequence of real numbers and x is considered a variable.

It is common to refer simply to a series, and modify by inﬁnite/power when clarity requires. We

almost always have m = 0 or 1, and it is common for examples to be centered at c = 0.

Example 2.6. Using the geometric series formula, we see that

∞

∑

n=0

( −1)

(x −4)

1 −

−(x−4)

x −2

whenever



−

x −4



< 1 ⇐⇒ 2 < x < 6

The series is valid (converges) only on a small subinterval of

the implied domain of the function x 7→

x−2

. The behavior

of both as x → 2

should not be a surprise; evaluating the

power series results in the divergent inﬁnite series

∑

1 = +∞

By contrast, as x → 6

−

, we see that limits and inﬁnite series

do not interact the way we might expect,

lim

x→6

−

∞

∑

n=0

( −1)

(x −4)

= lim

x→6

−

x −2

∞

∑

n=0

lim

x→6

−

( −1)

(x −4)

∑

( −1)

= DNE

with the last divergent by oscillation.

−6

−4

−2

−2 2 4 6

Power Series

f (x) =

x−2

As the example shows, we cannot take limits inside an inﬁnite sum; understanding when we can do

this is one of our primary goals.

lim inf



n+1



≤ lim inf

1/n

≤ lim sup

1/n

≤ lim sup



n+1



Radius and Interval of Convergence

At any real number x, a series may converge absolutely, converge conditionally, diverge to ±∞, or

diverge by oscillation. A series deﬁnes a function whose implied domain is the set on which the

series converges. In the previous example, the domain was an interval (2, 6). By applying the root

test (Theorem 2.3), we can show that this holds for every series.

Theorem 2.7 (Root Test for Power Series). Given a series

∞

∑

n=0

(x −c)

, deﬁne

R =

lim sup

1/n

Exactly one of the following is true:

R = ∞ the series converges absolutely for all x ∈ R

R = 0 the series converges only when x = c

R ∈ R

the series converges absolutely when

x −c

< R and diverges when

x −c

> R

Proof. For each ﬁxed x ∈ R, let b

= a

(x −c)

and apply the root test to

∑

, noting that

lim sup

1/n











0 if x = c or R = ∞

∞ if x = c and R = 0

lim sup

1/n

x −c

otherwise

In the ﬁnal situation, lim sup

1/n

< 1 ⇐⇒

x −c

< R, etc.

Deﬁnition 2.8. The radius of convergence is the value R deﬁned in Theorem 2.7. The interval of

convergence is the set of values x for which the series converges; the implied domain.

Radius of convergence Interval of convergence

∞ R = (−∞, ∞)

0 {c}

R ( c − R, c + R), (c − R, c + R], [c − R, c + R), or [c − R, c + R]

In the third case convergence/divergence at the endpoints of the interval of convergence must be

tested separately.

By applying Corollary 2.4, we obtain a more user-friendly result.

Corollary 2.9 (Ratio Test for Power Series). If the limit exists, R = lim

n→∞



n+1



Since

≥ 0, we here adopt the convention that

= ∞ and

∞

= 0. With similar caveats, it is also reasonable to write

R = lim inf

−1/n

Examples 2.10. 1. The series

∞

∑

n=1

is centered at 0. The ratio test tells us that

R = lim

n→∞



n+1



= lim

n→∞

1/n

1/(n + 1)

= lim

n→∞

n + 1

= 1

Test the endpoints of the interval of convergence separately:

x = 1

∑

= ∞ diverges

x = −1

∑

(−1)

converges (conditionally)

We conclude that the interval of convergence is [−1, 1).

It can be seen that the series converges to −ln(1 − x) on its

interval of convergence. As in Example 2.6, this function has

a larger domain (−∞, −1), than that of the series.

−1

−3 −2 −1 1

y =

∞

∑

n=0

y = −ln(1 − x)

2. The series

∞

∑

n=1

similarly has

R = lim

n→∞



n+1



= lim

n→∞

( n + 1)

= 1

Since

∑

is absolutely convergent, we conclude that the power series also converges abso-

lutely at x = ±1; the interval of convergence is [−1, 1].

3. The series

∞

∑

n=0

converges absolutely for all x ∈ R, since

R = lim

n→∞



n+1



= lim

n→∞

( n + 1)!

= lim

n→∞

( n + 1) = ∞

You should recall from elementary calculus that this series converges to the natural exponential

function exp(x) = e

everywhere on R; indeed this is one of the common deﬁnitions of the

exponential!

4. The series

∞

∑

n=0

n!x

has R = 0, and thus only converges at its center x = 0.

5. Let a





if n is even and





if n is odd. If we try to apply the ratio test to the series

∞

∑

n=0

, we see that



n+1



(





2n+1

if n even





2n+1

if n odd

=⇒ lim sup



n+1



= ∞ = 0 = lim inf



n+1



The ratio test therefore fails. However, by the root test,

1/n

(

if n even

if n odd

=⇒ R =

lim sup

1/n

3/2

It is easy to check that the series diverges at x = ±

; the interval of convergence is (−

With the help of the root test, we can understand the domain of a power series. The issues of limits,

continuity, differentiability and integrability are more delicate. We will return to these once we’ve

developed some of the ideas around convergence for sequences of functions.

Exercises 23 1. For each power series, ﬁnd the radius and interval of convergence:

(a)

∑

( −1)

(b)

∑

( n + 1)

(x −3)

(c)

∑

√

(d)

∑

√

(x + 7)

(e)

∑

(x −π)

(f)

∑

√

2n+1

2. For each n ∈ N let a



4+2(−1)



(a) Find lim sup

1/n

, lim inf

1/n

, lim sup



n+1



and lim inf



n+1



(b) Do the series

∑

and

∑

( −1)

converge? Why?

∑

3. Suppose that

∑

has radius of convergence R. If lim sup

> 0, prove that R ≤ 1.

4. On the interval (−

), express the series in Example 2.10.5 as a simple function.

(Hints: Use geometric series formulæ and the fact that the value of an absolutely convergent series is

independent of rearrangements)

5. Consider the power series

∞

∑

n=1

(x −7)

5n+1

(x −7) +

(x −7)

+ ···

Since only one in ﬁve of the terms are non-zero, it is a little tricky to analyze using a na

ıve

application of our standard tests.

(a) Explain why the ratio test for power series (Corollary 2.9) does not apply.

(b) Writing the series as

∑

(x −7)

, observe that







m−1

(m−1)

if m ≡ 1 mod 5

0 otherwise

Use the root test (Theorem 2.7) and your understanding of elementary limits to directly

compute the radius of convergence.

∑

(x −7)

5n+1

∑

. Apply the ratio test for inﬁnite series (Corol-

lary 2.4): what do you observe? Use your observation to compute the radius of conver-

gence of the original series in a simpler manner than part (a).

(d) Finally, check the endpoints to determine the interval of convergence.

24 Uniform Convergence

In this section we consider sequences ( f

) of functions f

: U → R.

Example 2.11. For each n ∈ N, consider f

: (0, 1) → R : x 7→ x

(x)

0 1

(x)

0 1

(x)

0 1

(x)

0 1

There turn out to be several good notions of convergence for sequences of functions; the simplest it

where, for each x, ( f

(x)) is treated as a separate sequence of real numbers.

Deﬁnition 2.12. Suppose a function f and a sequence of functions ( f

) are given. We say that ( f

)

converges pointwise to f on U if,

∀x ∈ U, lim

n→∞

(x) = f (x)

It is common to write ‘ f

→ f pointwise.’ For reference, we state two equivalent rephrasings:

1. ∀x ∈ U,

(x) − f (x)

−−−→

n→∞

2. ∀x ∈ U, ∀ϵ > 0, ∃N such that n > N =⇒

(x) − f (x)

< ϵ.

As we’ll see shortly, the relative positions of the quantiﬁers (∀x and ∃N) is crucial: in this deﬁnition,

the value of N is permitted to depend on x as well as ϵ.

Example (2.11, mk. II). The sequence ( f

) converges pointwise on the domain (0, 1) to

f : (0, 1) → R : x 7→ 0

We prove this explicitly as a sanity check. First observe that

(x) − f (x)

= x

Suppose x ∈ (0, 1), that ϵ > 0 is given, and let N =

ln ϵ

ln x

Then

n > N =⇒ n ln x < ln ϵ =⇒ x

< ϵ

where the inequality switches sign since ln x < 0.

(x)

0 1

···

The example is nice in that a sequence of continuous functions converges pointwise to a continuous

function. Unfortunately, this desirable situation is not universal.

Example (2.11, mk. III). Extend the domain to include

x = 1; deﬁne

: (0, 1] → R : x 7→ x

Each g

is a continuous function, however its pointwise limit

g(x) =

(

0 if x < 1

1 if x = 1

has a jump discontinuity at x = 1.

(x)

0 1

···

With the goal of having convergence of functions preserve continuity, we make a tighter deﬁnition.

Deﬁnition 2.13. ( f

) converges uniformly to f on U if either

1. sup

x∈U

(x) − f (x)

−−−→

n→∞

0, or,

2. ∀ϵ > 0, ∃N such that ∀x ∈ U, n > N =⇒

(x) − f (x)

< ϵ

A common notation is f

⇒ f , though we won’t use it.

2ǫ

f (x)

(x)

Whenever n > N, the graph of f

(x) must lie between those of f (x) ± ϵ.

We’ll show that statements 1 and 2 are equivalent momentarily. For the present, compare with the

corresponding statements for pointwise convergence:

• As with continuity versus uniform continuity, the distinction comes in the order of the quantiﬁers:

in uniform convergence, x is quantiﬁed after N and so the same N works for all x.

• Uniform convergence implies pointwise convergence.

For the last time, we revisit our main example.

Example (2.11, mk. IV). If f

: (0, 1) → R : x 7→ x

and f are deﬁned as before, then the pointwise

convergence f

→ f is non-uniform. We show this using both criteria.

1. For every n,

sup

x∈(0,1)

(x) − f (x)

= sup{x

: 0 < x < 1} = 1 ↛ 0

2. Suppose the convergence were uniform and let ϵ =

. Then

∃N ∈ N such that ∀x ∈ (0, 1), n > N =⇒ x

Since N ∈ N, a simple choice results in a contradiction;

x =





N+1

∈ (0, 1) =⇒ x

N+1

−ǫ

Theorem 2.14. The criteria for uniform convergence in Deﬁnition 2.13 are equivalent.

Proof. (1 ⇒ 2) This follows from the fact that

∀x ∈ U,

(x) − f (x)

≤ sup

x∈U

(x) − f (x)

(2 ⇒ 1) Suppose ϵ > 0 is given. Then

∃N ∈ R such that ∀x ∈ U, n > N =⇒

(x) − f (x)

But then

n > N =⇒ sup

x∈U

(x) − f (x)

≤

< ϵ

Somewhat amazingly, the subtle change of deﬁnition results in the preservation of continuity.

Theorem 2.15. Suppose that ( f

) is a sequence of continuous functions. If f

→ f uniformly, then

f is continuous.

Proof. We demonstrate the continuity of f at a ∈ U. Let ϵ > 0 be given.

• Since f

→ f uniformly,

∃N such that ∀x ∈ U, n > N =⇒

f (x) − f

(x)

• Choose any n > N. Since f

is continuous at a,

∃δ > 0 such that

x −a

< δ =⇒

(x) − f

(a)

(†)

Simply put these together with the triangle inequality to see that

x −a

< δ =⇒

f (x) − f (a)

≤

f (x) − f

(x)

(x) − f

(a)

(a) − f (a)

= ϵ

We need not have ﬁxed a at the start of the proof. Rewriting (†) to become

∃δ > 0 such that ∀x, a ∈ U,

x −a

< δ =⇒

(x) − f

(a)

proves a related result.

Corollary 2.16. If f

→ f uniformly where each f

is uniformly continuous, then f is uniformly

continuous.

Examples 2.17. 1. Let f

(x) = x +

. This is continuous on R for all x, and converges pointwise

to the continuous function f : x 7→ x.

(a) On any bounded interval [−M, M] the convergence f

→ f is uniform,

sup

x∈[−M,M]

(x) − f (x)

= sup



: x ∈ [−M, M]



−−−→

n→∞

(b) On any unbounded interval, R say, the convergence is non-uniform,

sup

x∈R

(x) − f (x)

= sup



: x ∈ R



= ∞

2. Consider f

(x) =

1+x

; this is continuous on

( −1, ∞) and converges pointwise to

f (x) =











0 if x > 1

if x = 1

1 if −1 < x < 1

We consider the convergence f

→ f on several

intervals.

(x)

−1 0 1 2 3

(a) On [2, ∞), the pointwise limit is continuous. Moreover, f

(x) is decreasing, whence

sup

x∈[2,∞)

(x) −0

1 + 2

−−−→

n→∞

and the convergence is uniform. Alternatively; if ϵ ∈ (0, 1), let N = log

( ϵ

−1

−1), then

∀x ≥ 2, n > N =⇒

(x) −0

1 + x

≤

1 + 2

= ϵ

The same argument shows that f

→ f uniformly on any interval [a, ∞) where a > 1.

(b) On [1, ∞) the convergence is not uniform, since the pointwise limit is discontinuous,

f (x) =

(

0 if x > 1

if x = 1

sup

x∈[1,∞)

(x) − f (x)

= sup



1 + x

: x > 1



−−−→

n→∞

(d) Similarly, for any a ∈ (0, 1), the convergence f

→ f is uniform on [0, a], this time to the

(continuous) constant function f (x) = 1,

sup

x∈[0,a]

(x) −1



1 −

1 + a



1 + a

−−−→

n→∞

(e) Finally, on (−1, 1) the convergence is not uniform,

sup

x∈[0,1)

(x) − f (x)

= sup



1 + x

: x ∈ [0, 1)



−−−→

n→∞

Exercises 24 1. For each sequence of functions deﬁned on [0, ∞):

(i) Find the pointwise limit f (x) as n → ∞.

(ii) Determine whether f

→ f uniformly on [0, 1].

(iii) Determine whether f

→ f uniformly on [1, ∞).

(a) f

(x) =

(b) f

(x) =

1 + x

(x) =

n + x

(d) f

(x) =

1 + nx

(e) f

(x) =

1 + nx

2. Let f

(x) =



x −



. If f (x) = x

, we clearly have f

→ f pointwise on any domain.

(a) Prove that the convergence is uniform on [−1, 1].

(b) Prove that the convergence is non-uniform on R.

3. For each sequence, ﬁnd the pointwise limit and decide if the convergence is uniform.

(a) f

(x) =

1+2 cos

(nx)

√

for x ∈ R.

(b) f

(x) = cos

(x) on [−π/2, π/2].

4. For each n ∈ N, consider the continuous function

: [0, 1] → R : x 7→ nx

(1 − x)

(a) Given 0 ≤ x < 1, let a ∈ (x, 1). Explain why ∃N such that

n > N =⇒

n+1

(x)

≤ a

(x)

Hence conclude that the pointwise limit of ( f

) is the zero function.

(b) Use elementary calculus ( f

′

(x) = 0 ⇐⇒ . . .) to prove that the maximum value of f

located at x

1+n

. Hence compute

sup

x∈[0,1]

(x) − f (x)

and use it to show that the convergence f

→ 0 is non-uniform.

This shows that the converse to Theorem 2.15 is false, even on a bounded interval: the continuous

sequence ( f

) converges non-uniformly to a continuous function. Sketches of several f

are below.

0 1

−1

y = f

(x)

0 1

−1

y = f

(x)

0 1

−1

y = f

(x)

0 1

−1

y = f

(x)

5. Explain where the proof of Theorem 2.15 fails if f

→ f non-uniformly.

25 More on Uniform Convergence

While we haven’t yet developed calculus, our familiarity with basic differentiation and integration

makes it natural to pause to consider the interaction of these operations with sequences of functions.

We also consider a Cauchy-criterion for uniform convergence, which leads to the useful Weierstraß

M-test.

Example 2.18. Recall that f

(x) = x

converges uniformly to f (x) = 0 on any interval [0, a] where

a < 1. We easily check that

(x) dx =

n + 1

n+1

−−−→

n→∞

0 =

f (x) dx

In fact the sequence of derivatives converge here also

(x) = nx

n−1

−−−→

n→∞

0 = f

′

(x)

It is perhaps surprising that integration interacts more nicely with uniform limits than does differen-

tiation. We therefore consider integration ﬁrst.

Theorem 2.19. Let f

→ f uniformly on [a, b] where the functions f

are integrable. Then f is

integrable on [a, b] and

lim

n→∞

(x)dx =

f (x)dx

Proof. Given ϵ > 0, note that

2(b−a)

dx =

. Since f

→ f uniformly, ∃N such that

∀x ∈ [a, b], n > N =⇒

(x) − f (x)

2(b − a)

=⇒ f

(x) −

2(b − a)

< f (x) < f

(x) +

2(b − a)

=⇒

(x) dx −

≤

f (x) dx ≤

(x) dx +

=⇒



(x) dx −

f (x) dx



≤

< ϵ

The appearance of uniform convergence in the proof is subtle. If N = N(ϵ) were allowed to depend

on x, then the integral

(x) dx would be meaningless: Which n would we consider? Larger than

N(x, ϵ) for which x? Taking n ‘larger’ than all the N(x, ϵ) might produce the absurdity n = ∞!

This assumes f is already integrable. Once we’ve properly deﬁned (Riemann) integrability at the end of the course, we

can insert the following

(x) dx −

≤ L( f ) ≤ U( f ) ≤

(x) dx +

=⇒ 0 ≤ U( f ) − L( f ) ≤ ϵ =⇒ U( f ) = L( f )

where U( f ) and L( f ) are the upper and lower Darboux integrals of f ; their equality shows that f is integrable on [a, b].

Examples 2.20. 1. Uniform convergence is not required for the integrals to converge as we’d like.

For instance, recall that extending the previous example to the domain [0, 1] results in non-

uniform convergence; however, we still have

(x) dx =

n + 1

−−−→

n→∞

0 =

f (x) dx

2. To obtain a sequence of functions f

→ f for which

↛

f requires a bit of creativity.

Consider the sequence

: [−1, 1] → R : x 7→

(

n −n

x if 0 < x <

0 otherwise

If 0 < x < 1, then for large n ∈ N we have

x ≥

=⇒ f

(x) = 0

−1 1

(x)

We conclude that f

→ 0 pointwise. Since the area under f

is a triangle with base

and height

n, the integral is constant and non-zero;

−1

(x) dx =

= 0 =

−1

f (x) dx

It should be obvious why the convergence f

→ 0 is non-uniform; why?

Derivatives and Uniform Limits We’ve already seen that a uniform limit of differentiable functions

might be differentiable (Example 2.18), but this shouldn’t be expected in general since even uniform

limits of differentiable functions can have corners!

Example 2.21. For each n ∈ N, consider the function

: [−1, 1] → R : x 7→

(

≥

• f

converges pointwise to f (x) =

• f

→ f uniformly since

sup

x∈[−1,1]

(x) − f (x)

→ 0

• Each f

is differentiable: f

′

(x) =











1 if x ≥

nx if

−1 if x ≤ −

• The uniform limit f is not differentiable at x = 0.

(x)

−1 0 1

−1

′

(x)

−1 1

−

Transferring differentiability to the limit of a sequence of functions is a bit messy.

Theorem 2.22. Suppose ( f

) is a sequence and g is a function on [a, b] for which:

• f

→ f pointwise;

• Each f

is differentiable with continuous derivative;

• f

′

→ g uniformly.

Then f

→ f uniformly on [a, b] and f is differentiable with derivative g.

The issue in the previous example is that the pointwise limit of the derived sequence ( f

′

) is discontin-

uous at x = 0 and therefore f

′

→ g isn’t uniform!

Proof. For any x ∈ [a, b], the fundamental theorem of calculus tells us that

′

( t) dt = f

(x) − f

(a)

By Theorem 2.19, the left side converges to

g(t) dt, while the right converges to f (x) − f (a). Since

′

→ g uniformly, we see that g is continuous and we can apply the fundamental theorem again:

g(t) dt = f (x) − f (a) is differentiable with derivative g.

The uniformity of the convergence f

→ f follows from Exercise 10.

Uniformly Cauchy Sequences and the Weierstraß M-Test

Recall that one may use Cauchy sequences to demonstrate convergence without knowing the limit in

advance. An analogous discussion is available for sequences of functions.

Deﬁnition 2.23. A sequence of functions ( f

) is uniformly Cauchy on U if

∀ϵ > 0, ∃N ∈ N such that ∀x ∈ U, m, n > N =⇒

(x) − f

(x)

< ϵ

Example 2.24. Let f

(x) =

∑

k=1

sin k

x be deﬁned on R. Given ϵ > 0, let N =

, then

m > n > N =⇒

(x) − f

(x)



∑

k=n+1

sin k



≤

∑

k=n+1

≤

∑

k=n+1

k(k −1)

∑

k=n+1

k −1

−

= ϵ

whence ( f

) is uniformly Cauchy.

Without the continuity assumption, the fundamental theorem of calculus doesn’t apply and the proof requires an

alternative approach. One can also weaken the hypotheses: if f

′

→ g uniformly and that ( f

(x)) converges for at least one

x ∈ [a, b], then there exists f such that f

→ f is uniform and f

′

= g.

As with sequences of real numbers, uniformly Cauchy sequences converge; in fact uniformly!

Theorem 2.25. A sequence ( f

) is uniformly Cauchy on U if and only if it converges uniformly to

some f : U → R.

Proof. (⇒) Let ( f

) be uniformly Cauchy on U. For each x ∈ U, the sequence ( f

(x)) ⊆ R is Cauchy

and thus convergent. Deﬁne f : U → R via

f (x) := lim

n→∞

(x)

We claim that f

→ f uniformly. Let ϵ > 0 be given, then

∃N ∈ N such that m > n > N =⇒

(x) − f

(x)

=⇒ f

(x) −

< f

(x) < f

(x) +

=⇒ f

(x) −

≤ f (x) ≤ f

(x) +

(take limits as m → ∞)

=⇒

(x) − f (x)

≤

< ϵ

(⇐) This is Exercise 2.

Example (2.24, mk. II). Since ( f

) is uniformly Cauchy on R, it converges uniformly to some

f : R → R. It seems reasonable to write

f (x) =

∞

∑

n=1

sin n

The graph of this function looks somewhat bizarre:

−1

f (x)

2ππ−2π −π

Since each f

is (uniformly) continuous, Theorem 2.15 says that f is also (uniformly) continuous. By

Theorem 2.19, f (x) is integrable, indeed

f (x) dx = lim

n→∞

∑

k=1

−k

−4

cos k



∞

∑

n=1

(cos n

a −cos n

which converges (comparison test) for all a, b. By contrast, the derived sequence

′

(x) =

∑

k=1

cos k

does not converge for any x since cos n

−−→

k→∞

0. We should thus expect (though we offer no proof)

that f is nowhere differentiable.

The example generalizes. Suppose (g

) is a sequence of functions on U and deﬁne the series

∑

(x)

as the pointwise limit of the sequence ( f

) of partial sums

∞

∑

k=k

(x) := lim

n→∞

(x) where f

(x) =

∑

k=k

(x)

whenever the limit exists. The series is said to converge uniformly whenever ( f

) does so. Theorems

2.15, 2.19 and 2.22 immediately translate.

Corollary 2.26. Let

∑

be a series of functions converging uniformly on U. Then:

1. If each g

is (uniformly) continuous then

∑

is (uniformly) continuous.

2. If each g

is integrable, then

∑

(x) dx =

∑

(x) dx.

3. If each g

is continuously differentiable, and the sequence of derived partial sums f

′

converges

uniformly, then

∑

is differentiable and

∑

(x) =

∑

′

(x).

As an application of the uniform Cauchy criterion, we obtain an easy test for uniform convergence.

Theorem 2.27 (Weierstraß M-test). Suppose (g

) is a sequence of functions on U. Moreover as-

sume:

1. (M

) is a non-negative sequence such that

∑

converges.

2. Each g

is bounded by M

; that is

(x)

≤ M

Then

∑

(x) converges uniformly on U.

Proof. Let f

(x) =

∑

k=k

(x) deﬁne the sequence of partial sums. Since

∑

converges, its sequence

of partial sums is Cauchy (the Cauchy criterion for inﬁnite series); given ϵ > 0,

∃N such that m > n > N =⇒

∑

k=n+1

< ϵ

However, by assumption,

m > n > N =⇒

(x) − f

(x)



∑

k=n+1

(x)



≤

∑

k=n+1

(x)

≤

∑

k=n+1

< ϵ

The sequence of partial sums is uniformly Cauchy and thus uniformly convergent.

Example 2.28. Given the series

∞

∑

n=1

1+cos

(nx)

sin(nx), we clearly have



1 + cos

( nx)

sin(nx)



≤

for all x ∈ R

Since

∑

converges, the M-test shows that the original series converges uniformly on R.

Exercises 25 1. For each n ∈ N, let f

(x) = nx

when x ∈ [0, 1) and f

(1) = 0.

(a) Prove that f

→ 0 pointwise on [0, 1].

(Hint: recall Exercise 24.4 if you’re not sure how to prove this)

(b) By considering the integrals

(x) dx show that f

→ 0 is not uniform.

2. Prove that if f

→ f uniformly, then the sequence ( f

) is uniformly Cauchy.

3. (a) Suppose ( f

) is a sequence of bounded functions on U and suppose that f

→ f converges

uniformly on U. Prove that f is bounded on U.

(b) Give an example of a sequence of bounded functions ( f

) converging pointwise to f on

[0, ∞), but for which f is unbounded.

4. The sequence deﬁned by f

(x) =

1+nx

(Exercise 24.1) converges uniformly on any closed

interval [a, b] where 0 < a < b.

(a) Check explicitly that

(x) dx →

f (x) dx, where f = lim f

(b) Is the same thing true for derivatives?

5. Let f

(x) = n

−1

sin n

x be deﬁned on R.

(a) Prove that f

converges uniformly on R.

(b) Check that

( t) dt converges for any x ∈ R.

′

) converge? Explain.

6. Use the M-test to prove that

∞

∑

n=1

deﬁnes a continuous function on [−1, 1].

7. Prove that

∞

∑

n=1

sin x

( n + 1)

converges uniformly to a continuous function on the interval [−2, 2].

8. Prove that if

∑

converges uniformly on a set U and if h is a bounded function on U, then

∑

converges uniformly on U.

(Warning: you cannot simply write

∑

= h

∑

)

9. Consider Example 2.20.2.

(a) Check explicitly that the convergence isn’t uniform by computing sup

x∈[−1,1]

(x) − f (x)

(b) Prove that f

→ 0 pointwise on (0, 1] using the ϵ–N deﬁnition of convergence: that is,

given ϵ > 0 and x ∈ (0, 1], ﬁnd an explicit N(x, ϵ) such that

n > N =⇒

f (x)

< ϵ

What happens to your choice of N(x, ϵ) as x → 0

10. Suppose ( f

′

) converges uniformly on [a, b] and that each f

′

is continuous.

(a) Use the fact that ( f

′

) is uniformly Cauchy to prove that ( f

) is uniformly Cauchy and thus

converges uniformly to some function f .

(Hint:

(x) − f

(x)



′

( t) − f

′

( t) dt



. . .)

(b) Explain why we need not have assumed the existence of f in Theorem 2.22.

26 Differentiation and Integration of Power Series

In this section we specialize our recent results to power series. While everything will be stated for

series centered at x = 0, all are easily translated to arbitrary centers.

Theorem 2.29. Let

∑

be a power series with radius of convergence R > 0 and let T ∈ (0, R).

Then:

1. The series converges uniformly on [−T, T].

2. The series is uniformly continuous on [−T, T] and continuous on (−R, R).

Proof. This is an easy application of the M-test. For each k, deﬁne M

T < R =⇒

∑

converges absolutely =⇒

∑

converges

By the M-test and Corollary 2.26, the power series converges uniformly on [−T, T] to a uniformly

continuous function.

Finally, every x ∈ (−R, R) lies in some such interval (take T =

), whence the power series is

continuous on (−R, R).

Example 2.30. On its interval of convergence (−1, 1), the geometric series

∞

∑

n=0

converges point-

wise to

1−x

; convergence is uniform on any interval [−T, T] ⊆ (−1, 1).

We needn’t use the Theorem for this is simple to verify directly: writing f , f

for the series and its

partial sums,

(x) − f (x)



1 − x

n+1

1 − x

−

1 − x



n+1

1 − x



=⇒ sup

x∈[−T,T]

(x) − f (x)

n+1

1 −T

−−−→

n→∞

By contrast, the convergence is non-uniform on (−1, 1);

sup

x∈(−1,1)

(x) − f (x)

= ∞

Theorem 2.31. Suppose

∞

∑

n=0

has radius of convergence R > 0. Then the series is integrable and

differentiable term-by-term on the interval (−R, R). Indeed for any x ∈ (−R, R),

∞

∑

n=0

∞

∑

n=1

n−1

and

∞

∑

n=0

dt =

∞

∑

n=0

n + 1

n+1

where both series also have radius of convergence R.

Proof. Let f (x) =

∑

have radius of convergence R, and observe that

lim sup

1/n

= lim n

1/n

lim sup

1/n

whence

∑

also has radius of convergence R. At any given non-zero x ∈ (−R, R), we may write

∞

∑

n=1

n−1

= x

−1

∞

∑

n=1

to see that the derived series also has radius of convergence R. On any interval [−T, T] ⊆ (−R, R), the

derived series converges uniformly (Theorem 2.29). Since each a

is continuously differentiable,

Corollary 2.26 says that f is differentiable on [−T, T] and that

′

(x) =

∞

∑

n=0

∞

∑

n=1

n−1

Since any x ∈ (−R, R) lies in some such interval [−T, T], we are done.

The corresponding result for integrals is Exercise 6.

We postpone the canonical examples until after the next result.

Continuity at Endpoints?

There is one small hole in our analysis. If a series has radius of convergence R we know that it

converges and is continuous on (−R, R). But what if it additionally converges at x = ±R? Is the

series continuous at the endpoints? The answer is an unequivocal yes, though this small beneﬁt

requires a lot of work!

Theorem 2.32 (Abel’s Theorem). Power series are continuous on their full interval of convergence.

Examples 2.33. 1. Apply our results to the geometric series;

(1 − x)

1 − x

∞

∑

n=1

n−1

∞

∑

n=0

( n + 1)x

= 1 + 2x + 3x

+ 4x

+ ···

ln( 1 −x) = −

1 −t

dt = −

∞

∑

n=0

n + 1

n+1

= −

∞

∑

n=1

= −



x +

+ ···



with both series valid on (−1, 1). In fact the ﬁrst series has the same interval of convergence,

while the second is [−1, 1). By Abel’s Theorem and the fact that logarithms are continuous, we

have equality at x = −1 and the famous identity

ln 2 =

∞

∑

n=1

( −1)

n+1

= 1 −

−

+ ···

This example shows that while the integrated and differentiated series have the same radius of

convergence as the original, convergence at the endpoints need not be the same in all cases.

2. Substitute x 7→ −x

in the geometric series and integrate term-by-term: if

< 1, then

1 + x

∞

∑

n=0

( −1)

=⇒ arctan x =

∞

∑

n=0

( −1)

2n + 1

2n+1

In fact the arctangent series also converges at x = ±1; Abel’s Theorem says it is continuous on

[−1, 1]. Since arctangent is continuous (on R!) we recover another famous identity

= arctan 1 =

∞

∑

n=0

( −1)

2n + 1

= 1 −

−

+ ···

As with the identity for ln 2, this is a very slowly converging alternating series and therefore

doesn’t provide an efﬁcient method for approximating π.

3. The series f (x) =

∞

∑

n=0

( −1)

(2n)!

has radius of convergence ∞. Differentiate to obtain

′

(x) =

∞

∑

n=1

( −1)

2n−1

(2n −1)!

∞

∑

n=0

( −1)

n+1

(2n + 1)!

2n+1

This series is also valid for all x ∈ R. Differentiating again,

′′

(x) =

∞

∑

n=0

( −1)

n+1

(2n)!

= −

∞

∑

n=0

( −1)

(2n)!

= −f (x)

Recalling that f (x) = cos x is the unique solution to the initial value problem

(

′′

(x) = −f (x)

f (0) = 1, f

′

(0) = 0

We conclude that, ∀x ∈ R,

cos x =

∞

∑

n=0

( −1)

(2n)!

sin x = −f

′

(x) =

∞

∑

n=0

( −1)

(2n + 1)!

2n+1

These expressions can instead be taken as the deﬁnitions of sine and cosine. As promised earlier in

the course, continuity and differentiability now come for free. One difﬁculty with this deﬁnition is

believing that it has anything to do with right-triangles!

We can similarly deﬁne other common transcendental functions using power series: for instance

exp(x) =

∞

∑

n=0

Example 2.33.1 could be taken as a deﬁnition of the logarithm on the interval ( 0, 2],

ln x = ln(1 − (1 −x)) = −

∞

∑

n=1

(1 − x)

∞

∑

n=1

( −1)

n+1

(x −1)

though this is unnecessary since it is more natural to deﬁne ln as the inverse of the exponential.

Proof of Abel’s Theorem (non-examinable)

This requires a lot of work, so feel free to omit on a ﬁrst reading!

First observe that there is nothing to check unless 0 < R < ∞. By the change of variable x 7→ ±

, it

is enough for us to prove the following:

∞

∑

n=0

convergent and f (x) =

∞

∑

n=0

on (−1, 1) =⇒ lim

x→1

−

f (x) =

∞

∑

n=0

Proof. Let s

∑

k=0

and write s = lim s

∑

. It is an easy exercise to check that

∑

k=0

= s

+ (1 − x)

n−1

∑

k=0

< 1, then (since s

→ s) lim s

= 0, whence we obtain

∀x ∈ (−1, 1), f (x) = (1 − x)

∞

∑

n=0

Let ϵ ∈ (0, 1) be given and ﬁx x ∈ (0, 1). Then

∃N ∈ N such that n > N =⇒

−s

(∗)

Use the geometric series formula

∞

∑

n=0

1−x

and write h(x) = (1 −x)



∑

n=0

( s

−s)x



to observe

f (x) − s



(1 − x)

∞

∑

n=0

−s



(1 − x)

∞

∑

n=0

−s(1 − x)

∞

∑

n=0



(1 − x)

∞

∑

n=0

( s

−s)x



= (1 − x)



∑

n=0

( s

−s)x

∞

∑

n=N+1

( s

−s)x



≤ (1 − x)



∑

n=0

( s

−s)x



+ (1 − x)



∞

∑

n=N+1

( s

−s)x



(△-inequality)

< h(x) +

(1 − x)



∞

∑

n=N+1



(by (∗))

≤ h(x) +

Since h > 0 is continuous and h(1) = 0, ∃δ > 0 such that x ∈ (1 − δ, 1) =⇒ h(x) <

(the

computation of a suitable δ is another exercise).

We conclude that lim

x→1

−

f (x) = s.

Exercises 26 1. (a) Prove that

∞

∑

n=1

(1 − x)

for

< 1.

(b) Evaluate

∞

∑

n=1

∞

∑

n=1

and

∞

∑

n=1

( −1)

2. (a) Starting with a power series centered at x = 0, evaluate the integral

1/2

1 + x

dx as an

inﬁnite series.

(b) (Harder) Repeat part (a) but for

1 + x

dx. What extra ingredients do you need?

3. The probability that a standard normally distributed random variable X lies in the interval [a, b]

is given by the integral

P(a ≤ X ≤ b) =

√

2π

exp



−



Find P(−1 ≤ X ≤ 1) as an inﬁnite series.

4. Deﬁne c(x) =

∞

∑

n=0

(2n)!

and s(x) =

∞

∑

n=0

2n+1

(2n + 1)!

(a) Prove that c

′

(x) = s(x) and that s

′

(x) = c(x).

(b) Prove that c(x)

−s(x)

= 1 for all x ∈ R.

(These functions are the hyperbolic sine and cosine: s(x) = sinh x and c(x) = cosh x)

5. Let a, b ∈ (−1, 1). Extending Example 2.30, show that the convergence

∑

1−x

is non-

uniform on any interval of the form (−1, a) or (b, 1).

6. Prove the integration part of Theorem 2.31.

7. Prove or disprove: If a series converges absolutely at the endpoints of its interval of convergence

then its convergence is uniform on the entire interval.

8. Complete the proof of Abel’s Theorem:

(a) Let s

∑

k=0

be the partial sum of the series

∑

. For each n, prove that,

∑

k=0

= s

+ (1 − x)

n−1

∑

k=0

(b) Suppose x > 0. Let S = max{

−s

: n ≤ N} and prove that h(x) ≤ S(1 − x

N+1

). Hence

ﬁnd an explicit δ that completes the ﬁnal step.

27 The Weierstraß Approximation Theorem

A major theme of analysis is approximation; for instance power series are an example of (uniform)

approximation by polynomials. It is reasonable to ask whether any function can be so approximated.

In 1885, Weierstraß answered a speciﬁc case in the afﬁrmative.

Theorem 2.34 (Weierstraß). If f : [a, b] → R is continuous, then there exists a sequence of polyno-

mials converging uniformly to f on [a, b].

Suitable polynomials can be deﬁned in various ways. By scaling the domain, it is enough to do this

on [a, b] = [0, 1] where perhaps the simplest approach is via the Bernstein Polynomials,

f (x) :=

∑

k=0









(1 − x)

n−k

(

)

k!(n−k)!

is the binomial coefﬁcient)

We omit the proof due to length; Weierstraß’ original argument was completely different. Instead we

compute a couple of examples and give an important interpretation/application.

Examples 2.35. 1. Suppose f (x) = 2x if x <

and f (x) = 1 otherwise.

f (x) = f (0) (1 −x) f (0) + f (1) x = x

f (x) = f (0) (1 −x)

+ 2 f (

)x(1 − x) + f (1)x

= 2x(1 − x) + x

= x(2 − x)

0 1

f (x) = f (0) (1 −x)

+ 3 f (

)x(1 − x)

+ 3 f (

(1 − x) + f (1)x

= 0(1 − x)

+ 2x(1 − x)

+ 3x

(1 − x) + x

= x(2 − x) = B

f (x)

f (x) = 0(1 − x)

+ 2x(1 − x)

+ 6x

(1 − x)

+ 4x

(1 − x) + x

= x(x

−2x

+ 2)

The Bernstein polynomials B

f (x), B

f (x) and B

f (x) are drawn.

2. Now assume f (x) = x if x <

and 1 − x otherwise.

f (x) = f (0) (1 −x) + f (1)x = 0

f (x) = x(1 − x)

f (x) = 0(1 − x)

+ x(1 − x)

+ x

(1 − x) + 0x

= x(1 − x) = B

f (x)

0 1

f (x) = f (0) (1 −x)

+ f (

)·4x(1 − x)

+ f (

)·6x

(1 − x)

+ f (

)·4x

(1 − x) + f (1)x

= x(1 − x)

+ 3x

(1 − x)

+ x

(1 − x)

= x(1 − x)(1 + x − x

)

B´ezier curves (just for fun!)

The Bernstein polynomials arise naturally when con-

sidering B´ezier curves. These have many applications,

particularly in computer graphics. Given three points

A, B, C, deﬁne points on the line segments

−→

AB and

−→

for each t ∈ [0, 1], via

−→

AB(t) = (1 −t)A + tB

−→

BC(t) = (1 −t)B + tC

These points move at a constant speed along the cor-

responding segments. Now consider a point on the

moving segment between the points deﬁned above:

0 1

−→

B C

R(t) := (1 −t)

−→

AB(t) + t

−→

BC(t) = (1 − t)

A + 2t(1 −t)B + t

This is the quadratic B´ezier curve with control points A, B, C. The 2

Bernstein polynomial for a function

f is simply the quadratic B

ezier curve with control points

(

0, f (0)

)



, f (

)



and

(

1, f (1)

)

. The

picture

above shows B

f (x) for the above example.

We can repeat the construction with more control points: with four points A, B, C, D, one constructs

−→

AB(t),

−→

BC(t),

−→

CD(t), then the second-order points between these, and ﬁnally the cubic B

ezier curve

R(t) : = (1 − t)



(1 −t)

−→

AB(t) + t

−→

BC(t)



+ t



(1 −t)

−→

BC(t) + t

−→

CD(t)



= (1 −t)

A + 3t(1 −t)

B + 3t

(1 −t)C + t

where we now recognize the relationship to the 3

Bernstein polynomial.

The pictures show cubic B

ezier curves: the ﬁrst is the graph of the Bernstein polynomial

f (x) = 0(1 − x)

+ 3x(1 − x)

+ 3x

(1 − x) +

while the second is for the four given control points A, B, C, D.

To see these pictures move, visit https://www.math.uci.edu/

ndonalds/math140b/bezier.html

Exercises 27 1. Show that the closed bounded interval assumption in the approximation theorem

is required by giving an example of a continuous function f : ( −1, 1) → R which is not the

uniform limit of a sequence of polynomials.

2. If g : [a, b] → R is continuous, then f (x) := g



( b − a)x + a



is continuous on [0, 1]. If P

→ f

uniformly on [0, 1], prove that Q

→ g uniformly on [a, b], where

(x) = P



x −a

b − a



3. Use the binomial theorem to check that every Bernstein polynomial for f (x) = x is B

f (x) = x

itself!

4. Find a parametrization of the cubic B

ezier curve with control points (1, 0), (0, 1), (−1, 0) and

(0, −1). Now sketch the curve.

(Use a computer algebra package if you like!)

5. (Hard) Show that the Bernstein polynomials for f (x) = x

are given by

f (x) =

x +

n −1

and thus verify explicitly that B

f → f uniformly.