2 Sequences and Series of Functions

If ( f

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

? This question is of huge 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 easily differentiated and integrated using the power law. What, however, are we to make of

the series

f (x) :=

∞

∑

n=0

= 1 + x + x

+ ··· ?

Does this make sense as a function? What is its domain? Does it equal the limit of the sequence

( f

) in any meaningful way? Is it continuous, differentiable, integrable? If so, can we compute its

derivative or integral term-by-term: for instance, is it legitimate to write

′

(x) =

∞

∑

n=1

n−1

= 1 + 2x + 3x

+ ··· ?

To many in Newton’s time, such technical questions were less important than the application of cal-

culus to the natural sciences. For the 18

and 19

century mathematicians who followed, however,

the widespread application of calculus only increased the imperative to rigorously address these

issues.

2.23 Power Series

First we review some of the important deﬁnitions, examples and results concerning 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

∑

is said to converge, diverge to inﬁnity or diverge by oscillation

as does (s

∑

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 compli-

cated series. Make sure they are familiar!

1. (Geometric series) If r is 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 or refer 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. By partial fractions, 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)

< 1 +

∞

∑

n=1

n(n −1)

= 2

Since (s

) is monotone-up and 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

is monotone-up, 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 (essentially the proof of the alternating series test).

Root Test: β < 1 =⇒ ∃ϵ > 0 such that

1/n

≤ 1 −ϵ (for large n) =⇒

∑

converges by comparison with

∑

(1 −ϵ)

β > 1 =⇒ some subsequence of (

1/n

) converges to β > 1 =⇒ b

↛ 0 =⇒

∑

diverges (n

-term test).

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

yield a test that is often easier to

apply.

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 are now ready to properly deﬁne and analyze our main objects of interest.

Deﬁnition 2.5. A power series centered at c ∈ R with coefﬁcients a

∈ R is a formal expression

∞

∑

n=m

(x −c)

where x ∈ R is considered a variable. A power series is a function whose implied domain is the set

of x for which the resulting inﬁnite series converges.

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

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

Example 2.6. By the geometric series formula,

∞

∑

n=0

( −1)

(x −4)

1 −

−(x−4)

x −2

whenever



−

x −4



< 1 ⇐⇒ 2 < x < 6

The series is valid (converges) only on the subinterval (2, 6) of

the implied domain of the function x 7→

x−2

The behavior as x → 2

is unsurprising, since 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 as 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 series being divergent by oscillation.

−6

−4

−2

−2 2 4 6

Power Series

f (x) =

x−2

The example shows that we cannot blindly take limits inside an inﬁnite sum; understanding precisely

when this is possible is one of our primary goals.

You should have encountered these previously: lim inf



n+1



≤ lim inf

1/n

≤ lim sup

1/n

≤ lim sup



n+1



Radius and Interval of Convergence

The implied domain of the series in Example 2.6 turned out to be an interval (2, 6). Somewhat amaz-

ingly, the root test (Theorem 2.3) shows that the same is true for every power series!

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

∑

(x −c)

, deﬁne

R =

lim sup

1/n

The precisely one of the following statements holds:

R ∈ (0, ∞) the series converges absolutely when

x −c

< R and diverges when

x −c

> R

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

R = 0 the series converges only at the center x = c

Proof. For each ﬁxed x ∈ R, let b

= a

(x −c)

and apply the root test to

∑

, noting that

lim sup

1/n











lim sup

1/n

x −c

if R ∈ (0, ∞)

0 if R = ∞ or x = c

∞ if R = 0 and x = c

In the ﬁrst 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 conver-

gence is the set of x ∈ R for which the series converges; its implied domain.

Radius of convergence Interval of convergence

R = 0, ∞ ( c − R, c + R), (c − R, c + R], [c − R, c + R) or [c − R, c + R]

∞ R = (−∞, ∞)

0 {c}

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

tested for separately.

The ratio test (Corollary 2.4) provides a more user-friendly version.

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

n→∞



n+1



The ratio test is weaker than the root test: as Example 2.10.5 shows, there exist series for which the

ratio test is be inconclusive.

Since

≥ 0, we here adopt the conventions

= ∞,

∞

= 0. With similar caveats, one can write R = lim inf

−1/n

Since every sequence has a limit superior, this really is a deﬁnition. Whether one can easily compute R is another matter. . .

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 (later) 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 function.

4. The series

∑

∞

n=0

n!x

has R = lim

(n+1)!

= 0. It therefore converges only 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 is therefore inconclusive. 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 the domain of a power series is fully understood. 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 2.23. Key concepts: Power Series, Radius/interval of convergence R =

lim sup

1/n

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) Does the series

∑

converge? What about

∑

( −1)

? 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.

2.24 Uniform Convergence

In this section we consider sequences ( f

) of functions f

: U → R and their limits.

Example 2.11. For each n ∈ N, deﬁne f

: (0, 1) → R : x 7→ x

. Several examples are graphed.

(x)

0 1

(x)

0 1

(x)

0 1

(x)

0 1

There are several useful notions of convergence for sequences of functions. The simplest is where,

for each input x, ( f

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

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

) are given, all with domain

U. 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, here are two equivalent rephrasings:

1. ∀x ∈ U, lim

n→∞

(x) − f (x)

= 0;

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

(x) − f (x)

< ϵ.

As we’ll see shortly, the relative position of the quantiﬁers (∀x, ∃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 U = (0, 1) to

f : (0, 1) → R : x 7→ 0

As a sanity check, we prove this explicitly. 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). 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

···

We’d like the limit of a sequence of continuous functions to itself be continuous. With this goal in

mind, 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)

As pictured, whenever n > N, the graph of f

(x) must lie between 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 locations x.

• Uniform convergence implies pointwise convergence.

Example (2.11, mk. IV). For the ﬁnal time we revisit our main example. If f

(x) = x

and f (x) = 0

are deﬁned on U = (0, 1), then f

→ f non-uniformly. We show this using both criteria.

1. For every n,

sup

x∈(0,1)

(x) − f (x)

= sup{x

: 0 < x < 1} = 1 −→ 0

which plainly fails to converge to zero.

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 immediately 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

Amazingly, this subtle change of deﬁnition is enough to preserve continuity.

Theorem 2.15. Suppose ( 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)

(†)

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. Suppose ( f

) is a sequence of uniformly continuous functions. If f

→ f uniformly,

then f is also 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 2.24. Key concepts: Pointwise & Uniform Convergence, Uniform conv preserves continuity

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.

2.25 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 concepts 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/Darboux) 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 ; 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 that 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). As the next example shows, this should not be expected in

general, since even uniform limits of differentiable functions can have corners.

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

: R → R : x 7→

(

≥

• Each f

is differentiable: f

′

(x) =











1 if x ≥

nx if

−1 if x ≤ −

• f

converges pointwise to f (x) =

, which is non-

differentiable at x = 0.

• f

→ f uniformly since

sup

x∈[−1,1]

(x) − f (x)

= f

(0) =

→ 0

(x)

−1 0 1

−1

′

(x)

−1 1

−

If our goal is to transfer differentiability to the limit of a sequence of functions, then we have some

work to do.

Theorem 2.22. Suppose ( f

) is a sequence and f , g functions, all with domain [a, b]. Suppose also:

• 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 (part II) tells us that

′

( t) dt = f

(x) − f

(a)

As n → ∞, Theorem 2.19 says the left side converges to

g(t) dt and the right to f (x) − f (a) (both

pointwise). Since f

′

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

theorem (part I):

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 this 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 ( 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 to be the pointwise limit:

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 lim

n→∞

cos n

x = 0. We should therefore 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 assume:

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 2.25. Key concepts: Uniform convergence preseves integration, Uniform Cauchyness, M-test

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 2.24.4 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 2.24.1) 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.

2.26 Differentiation and Integration of Power Series

We now 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 a straightforward application of the Weierstraß M-test (Theorem 2.27). For each k,

deﬁne M

, and observe that

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 pointwise

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), since

sup

x∈(−1,1)

(x) − f (x)

= ∞

Theorem 2.31. Suppose a power series

∑

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.

Exercise 7 discusses the corresponding result for integration.

We postpone the canonical examples until after the next result.

Continuity at Endpoints?

There is one small hole in our analysis. A series

∑

with radius of convergence R converges and

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

endpoints? The answer is yes, though demonstrating 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 +

+ ···



where both are valid on (−1, 1). In fact the ﬁrst series has exactly this interval of convergence,

whereas the second has [−1, 1). By Abel’s Theorem and the fact that logarithms are continuous,

we have equality at x = −1 and recover the famous identity

ln 2 =

∞

∑

n=1

( −1)

n+1

= 1 −

−

+ ···

This also shows that while integrated and differentiated series have the same radius of conver-

gence as the original, convergence at the endpoints need not be the same.

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 last expressions can be taken as the deﬁnitions of sine and cosine. As promised earlier, con-

tinuity and differentiability of these functions now come for free! The only real downside of 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 also 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 computa-

tion of a suitable δ is another exercise).

We conclude that lim

x→1

−

f (x) = s.

Exercises 2.26. Key concepts: Power series continuous on (−R, R), uniformly on any [−T, T] ⊂ (−R, R),

Power series differentiable and integrable term-by-term on (−R, R)

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. If f (x) =

∑

∞

n=0

(−1)

(2n)!

is deﬁned as in Example 2.33.3, prove that



f (x)





′

(x)



= 1.

What does (the converse of) Pythagoras’ Theorem say about f (x), at least when both it and

′

(x) are positive?

(Hint: Differentiate and evaluate at zero!)

5. 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)

6. 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).

7. Prove the integration part of Theorem 2.31.

8. 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.

9. 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.

2.27 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 f (x) = 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

−→

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.

Click on any of the pictures to see all of them move.

Exercises 2.27. Key concepts: Every continuous function is the uniform limit of a polynomial sequence

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.