CHAPTER 2. CAUCHY’S CRITERION
2.1. Roughly speaking, limn an = a, means that an are close to a for large n. This
implies the following property of the sequence {an }: for large n, an are close to each other.
Technically, a sequence with this property is called a Cauchy sequence.
Definition (ad hoc). A sequence {an } of real numbers is called a Cauchy sequence
if there is a sequence {αn } of positive numbers decreasing to zero such that, for all positive
integers m, n with m ≥ n, we have |am − an | ≤ αn .
The remark made at the beginning, stated precisely, is:
Observation A. A convergent sequence is a Cauchy sequence.
Proof: Suppose that {an } is a convergent sequence and a is its limit. Then there is
a sequence of positive numbers αn decreasing to zero such that |an − a| < αn for all n.
Now, for all m and n with m ≥ n, |am − an | = |am − a + a − an | ≤ |am − a| + |a − an | <
αm + αn ≤ 2αn . From the fact that {2αn }n is a sequence decreasing to zero, we see that
{an } is a Cauchy sequence. Q.E.D.
Remarkably, the converse of Observation A is also true:
THEOREM (Cauchy’s Criterion). A Cauchy sequence of reals converges.
Not only beautiful, this theorem is also extremely powerful. Often, by means of this
theorem, we can tell if a given sequence or series is convergent without actually knowing
its limit, which may be very difficult or even impossible to find.
2.2. The proof of this theorem involves sophisticated arguments and in-depth investigation of properties of the real number system, from which there are many things for us
to learn. It is advisable to proceed slowly. Let us make a few observations to get some
ideas of the proof.
Observation B. A Cauchy sequence {an } is bounded, that is, there exists some
M > 0 such that |an | ≤ M for all n.
Proof: By assumption, there exists a sequence of positive numbers αn decreasing to
zero such that |an − am | < αn for all m, n with m ≥ n. In particular, |am − a1 | ≤ α1 holds
1
for all m. So |am | = |am − a1 + a1 | ≤ |am − a1 | + |a1 | < α1 + |a1 |. Thus |an | ≤ M for all
n, where M = |a1 | + α1 . Q.E.D.
2.3. Next we introduce a key concept for establishing Cauchy’s criterion: cluster
points of a sequence.
Definition. Given a sequence {an } in R, a point c ∈ R is called a cluster point of
this sequence if, for each ε > 0, there are infinitely many n for which an ∈ (c − ε, c + ε).
The open interval (c − ε, c + ε) may be called a neighbourhood of c (with ε as its
radius, which indicates its size.) Thus, c is a cluster point of sequence if every neigbourhood
(no matter how small it is) of c contains infinitely many terms of this sequence. It is easy
to check that, if a sequence converges, then its limit is the unique cluster point of this
sequence; see Exercises 1 and 7. But, in general, a sequence may have many cluster points,
or none. For example, both +1 and −1 are cluster points of the sequence {(−1)n }n≥1 ,
while the sequence {n2 }n≥1 has no cluster points. The relevance of cluster points to the
proof of Cauchy’s criterion is indicated in the following observation:
Observation C. If c is a cluster point of a Cauchy sequence {an }n , then this Cauchy
sequence converges to c.
Proof: By assumption, there is a sequence of positive numbers αn converging to zero
such that |an − am | < αn for all m, n with m > n. Take an arbitrary positive integer n and
fix it for the moment. Since c is a cluster point of the given sequence, there are infinitely
many terms of this sequence lying in the neighbourhood (c − αn , c + αn ). So there must
be some m with m > n such that am ∈ (c − αn , c + αn ), that is, |am − c| < αn . Thus,
|an − c| ≤ |an − am | + |am − c| < αn + αn = 2αn .
We have shown |an − c| < 2αn for each n, implying limn an = c. Q.E.D.
2.4. This observation tells us that, in order to establish Cauchy’s criterion, all we
have to do is to show that a Cauchy sequence has a cluster point. We know that a Cauchy
sequence is bounded (Observation B). Hence it is enough to establish the following:
THEOREM (Bolzano–Weierstrass Theorem) Every bounded sequence of real
numbers has a cluster point.
Before starting to prove this theorem, let us shortly summarize what we have done so far.
We consider the following properties of a given sequence {an } of real numbers:
2
(a) {an } is a convergent sequence.
(b)
{an } is a Cauchy sequence.
(c)
{an } is a bounded sequence.
(d)
{an } has a cluster point.
We have established the implications (a) ⇒ (b) and (b) ⇒ (c). The implication (c) ⇒
(d) is the content of the Bolzano-Weierstrass theorem, which is the key for establishing
Cauchy’s criterion, as we have observed that (b),(d) ⇒ (a).
2.5. Now we try to prove the Bolzano–Weierstrass Theorem. So let {an } be a bounded
sequence, say |an | ≤ M for all n, where M is some positive number. We claim that we can
find a cluster point somewhere within the closed interval [−M, M ]. It is hopeless to look
into the set [−M, M ] point by point to search for a cluster point, because [−M, M ] has
infinitely many points; (actually the situation is much worse: [−M, M ] has uncountably
many points.) This situation is a bit like that a crime is committed and we are unable to
catch the criminal due to the fact that there are too many places for the criminal to hide.
Certainly we can argue that the criminal exists by “contradiction”: otherwise the crime
would never have occurred. Similarly, it is natural to prove Bolzano - Weierstrass theorem
by the method of contradiction. Let us proceed that way.
Proof (of Bolzano–Weierstrass Theorem): Let {an } be a bounded sequence, say |an | ≤
M for all n, where M > 0. We are going to show that {an } has a cluster point in [−M, M ].
Assume the contrary that none of the point in [−M, M ] is a cluster point of this sequence.
Take an arbitrary point b in [−M, M ]. Since b is not a cluster point of the sequence, it has
a neighbourhood, say Nb = (b − εb , b + εb ), containing at most finitely many terms of the
sequence, that is, the set {n : an ∈ Nb } is finite. Now {Nb }b∈
[−M,M ]
is a family of open
intervals covering [−M, M ], i.e.
[−M, M ] ⊆
b∈ [−M,M ]
Nb .
We claim that this covering family can be reduced to a finite subcover, that is, there is
a finite subset F of [−M, M ] such that [−M, M ] ⊆ b∈ F Nb . Let us accept this claim
for the moment and see how to finish the proof. For convenience, let us “spell out” F as
F = {b1 , b2 , · · · , br }. Then each of the open intervals Nb1 , Nb2 , . . . , Nbr contains at most
finitely many terms of {an }. So their union N ≡ Nb1 ∪ Nb2 ∪ · · · ∪ Nbr also contains at
most finitely many terms of {an }. But the whole sequence {an } is in the closed interval
3
[−M, M ], while [−M, M ] is contained in this union N . Clearly this is impossible. Thus
we have arrived at a contradiction.
2.6. It remains to prove our claim, which follows from the following:
LEMMA (Borel’s Covering Lemma). If C is a closed interval and if {Iα }α∈ A is
a family of open intervals which covers C, i.e. C ⊆ α∈ A Iα , then {Iα } can be reduced to
a finite subcover in the sense that there is a finite subset F of A such that C ⊆ α∈ F Iα .
The proof of this lemma needs some ingenuity to accomplish. Let me first explain
the idea behind this proof. Instead of looking at the single fixed interval [a, b], we look
at [a, r], where the right end point r varies from a to b. Observe that if r is close enough
to a, then [a, r] can be covered by a single Iα . Also observe that if [a, r] can be covered
by finitely many Iα , then r may “creep up” a bit, say, to r + δ with δ > 0, such that the
larger interval [a, r + δ] is still cover by finitely many Iα . Thus we would like to see that r
can move up from a to b so that at every stage [a, r] can be covered by finitely Iα . This is
just the rough idea, which has many frauds. Now we present the rigorous proof.
2.7. Proof of the covering lemma. Let S be the set of all those points r in [a, b]
such that [a, r] can be covered by finitely many Iα , i.e.
S = {r ∈ [a, b] : there is a finite subset F of A such that [a, r] ⊆
α∈ F
Iα }.
(When r = a, [a, r] is interpreted as the singleton set {a}.) Our goal is to show that b ∈ S.
Clearly a ∈ S, which guarantees that S is nonempty. Also, S is bounded above, because
r ≤ b for all r ∈ S. Hence, by the order completeness of R, the least upper bound for S
exists. Let s0 = sup S. Notice that, since S ⊆ [a, b], we have s0 ∈ [a, b].
Next we show that s0 is also in S. Since s0 ∈ [a, b] and since the family {Iα }α∈
A
covers [a, b], there exists some γ ∈ A such that s0 ∈ Iγ . Since Iγ is an open interval, there
exists some ε > 0 such that [s0 − ε, s0 ] ⊆ Iγ . As s0 = sup S, there exists some s1 ∈ S such
that s1 > s0 − ε.
4
α∈
Since s1 ∈ S, the interval [a, s1 ] can be covered by finitely many Iα , say [a, s1 ] ⊆
F Iα , where F is a finite subset of A. Now it is clear that [a, s0 ] ⊆
α∈ F0 Iα , where
F0 = F ∪ {γ}, a finite subset of A. Hence s0 ∈ S.
It remains to show that s0 = b. We have just shown that s0 ∈ S, i.e. [a, s0 ] is covered
by {Iα }α∈
F0
for some finite subset F0 of A. In particular, s0 ∈ Iβ for some β ∈ F0 . Since
Iβ is an open interval, we can find some δ > 0 such that [s0 , s0 + δ] ⊆ Iβ . Thus [a, s0 + δ]
is also covered by the finite subfamily {Iα }α∈
F0 .
If s0 < b did occur, then we could choose
δ > 0 so small that s0 + δ < b, which would imply s0 + δ ∈ S, contradicting the fact that
s0 is an upper bound of S. Therefore b = s0 ∈ S.
Q.E.D.
An alternative proof of the Bolzano-Weierstrass Theorem will be given in §2.10.
2.8. We have seen the expression “for infinitely many n” in the definition of cluster
points of sequences. There is a counterpart of this expression: “for all except finitely many
n”. We take a closer look of these two expressions in a general setting.
Let X be an arbitrary space, let {an }n≥1 be a sequence in this space, and let A be a
subset of X. We say that {an } is frequently in A if an ∈ A for infinitely many n. We
say that {an } is eventually in A if an ∈ A for all except finitely many n.
Thus, that {an } is frequently in A means that there is sequence of strictly increasing
positive integers k1 < k2 < k3 < · · · such that akn ∈ A for all n; in other words, {an } has
a subsequence lying in A. Another way to say: {an } is frequently in A if and only if, for
each N , there exists some N with n ≥ N such that an ∈ A. On the other hand, that {an }
is eventually in A means that there exists some N such that an ∈ A for all n ≥ N . To
summarize,
{an } is frequently in A
iff
∀N ∃n : n ≥ N and an ∈ A.
{an } is eventually in A
iff
∃N ∀n : n ≥ N ⇒ an ∈ A.
[Here “iff” is an abbreviation for “if and only if” invented by Paul Halmos.] It is clear that
if {an } is eventually in A, then it is frequently in A. The converse is obviously false: it is
an easy matter to write down a sequence which is frequently but not eventually in a set.
You should convince yourself that
{an } is not frequently in A
iff
{an } is eventually in X\A.
{an } is not eventually in A
iff
{an } is frequently in X\A.
5
2.9. Recall that a point c ∈ R is a cluster point of a sequence {an } if {an } is frequently
in (c − ε, c + ε) for every ε > 0. If we change the words “frequently in” by “eventually in”
in the last condition, we get the standard definition of limits:
(C) {an } converges to c iff {an } is eventually in (c − ε, c + ε) for every ε > 0.
Proof of (C). Suppose that limn an = c. According to our “ad hoc” definition of
limits, there is a sequence {αn } decreasing to zero such that |an − c| < αn for all n.
Let ε > 0 be given. Take a large enough N such that αN < ε. For n > N, we have
|an − c| < αn ≤ αN < ε. This shows that {an } is eventually in (c − ε, c + ε).
Conversely, suppose that {an } is eventually in (c − ε, c + ε) for every ε > 0. For each
positive integer n, let αn be the least upper bound of the set {|ak − c| : k ≥ n}. From
the way αn is defined, it is clear that the sequence {αn } is decreasing and |an − c| ≤ αn
holds for all n. Let ε > 0 be given. Then {an } is eventually in (c − ε, c + ε), that is, there
exists some N such that an is in (c − ε, c + ε) for all n ≥ N . Thus, |an − c| < ε for all
n ≥ N , from which it follows that αN ≡ sup{|an − c| : n ≥ N } ≤ ε. This shows that {αn }
decreases to zero. Q.E.D.
Recall once again that “{an } is eventually in (c − ε, c + ε)” means that there is N such
that an ∈ (c − ε, c + ε) for all n ≥ N . So
{an } converges to c
iff
∀ε > 0 : {an } is eventually in (c − ε, c + ε)
iff
∀ε > 0 ∃N ∀n : n ≥ N ⇒ |an − c| < ε.
The last description is usually taken to be the standard definition of limits of sequences:
Standard Definition of Limits for Sequences. A sequence {an } converges to c
if, for all given ε > 0, there exists some N such that |an − c| < ε for all n > N .
2.10. Recall the Bolzano–Weierstrass theorem: every bounded sequence of real numbers has a cluster point. Now we show that, among all cluster points of a bounded sequence,
there exist the largest one and the smallest one. (Our argument here actually gives an
alternative proof of Bolzano-Weierstrass theorem.)
So let {an } be the given sequence in R which is bounded, say −M ≤ an ≤ M for
all n. For each n, let bn be the supremum of the set Sn ≡ {an , an+
1 , an+ 2 , . . .},
that is,
bn = supk≥n ak . Clearly S1 ⊇ S2 ⊇ S3 ⊇ · · ·, from which it follows that b1 ≥ b2 ≥ b3 ≥ · · ·.
Also we have bn ≥ −M for all n. Thus {bn } is decreasing and bounded below. By the
6
proposition in §1.9, we know that bn converges to inf n≥1 bn . The limit limn bn ≡ inf n bn ,
is called the “lim sup” of {an } and is denoted either by lim supn an or limn an :
lim supn an ≡ limn an = inf n≥1 supk≥n ak .
We claim that the “lim sup” L = limn an has the following properties:
(1) For each ε > 0, {an } is eventually less than L + ε.
(2) For each ε > 0, {an } is frequently in (L − ε, L + ε).
Notice that (2) tells us that L is a cluster point and (1) tells us that there is no cluster
point strictly larger than L. Thus L is the largest cluster point of {an }.
Proof of (1): Since L is the limit of the sequence {bn }, (where bn = supk≥n ak ) there
exists some N such that bN < L+ε. Since bN is an upper bound for {aN , aN +
1 , aN+ 2 , . . .},
we have an < L + ε for all n ≥ N .
Proof of (2): Let N be an arbitrary positive integer. We have to find some n > N
such that an ∈ (L − ε, L + ε). Since L is the limit of the decreasing sequence {bn }n , we
can find some m > N such that L ≤ bm ≤ L + ε. Since L − ε is strictly less than bm and
bm is the supremum of the set {am , am+
1 , . . .},
we can find some n such that n ≥ m and
an > L − ε. Thus L − ε < an ≤ bm < L + ε, which tells us an ∈ (L − ε, L + ε).
In the same way, we can define the “lim inf” of {an }n by lim inf n an ≡ limn an =
supn≥1 inf k≥n ak , which is the smallest cluster point of {an }. For a bounded sequence
{an } in R, the limit limn an exists if and only if the “lim sup” and the “lim inf” of this
sequence coincide; see Exercise 5. Although this fact is not very deep, sometimes it helps
us to improve our presentation in dealing with convergence; see the next example.
2.11. Example: e, the natural base for logarithm. There are two common ways
1
to define e ≈ 2.17: as the sum of a series, e = ∞n= 0 n!
, and, as the limit of a sequence,
e = limn→
∞
(1 + n1 )n . We have to justify these two definitions by showing the convergence
of the above series and sequence, and also showing that their limits are the same. It is
1
relatively easy to see the convergence of the series
n! . All we need to check is that its
n 1
partial sums sn = k= 0 k! are bounded above. Indeed, for all n > 5,
sn ≡ 1 +
1
1
1
1
1
1
1
1
1
+ + + + ··· +
≤ 1+1+
+
+
+ ··· +
≤ 3,
1! 2! 3! 4!
n!
1.2 2.3 3.4
(n − 1)n
7
by “telescoping”. For the time being, let us denote the sum of the series
next task is to show that an ≡ (1 +
an =
n
k= 0
n!
k!(n − k)!
1 n
n)
1
n!
by S. Our
converges to S. The binomial theorem gives
k n
1
=
k=
n
0
1 n(n − 1) · · · (n − k + 1)
.
k!
nk
n−k+ 1
Now n(n−1) · · · (n−k +1)/nk = nn n−1
< 1. Thus an ≤ sn , (recall that sn is the
n ···
n
1
nth partial sum of
n! .) So lim sup an ≤ lim sn = S. It remains to show lim inf an ≥ S.
(From this we will get S ≤ lim inf an ≤ lim sup an ≤ S and hence lim an = S.)
To this end, we use the following trick. Fix a large positive integer N and cut off the
binomial expansion of an at N . Let n → ∞. This will tell us that lim inf an ≥ sN . Now
let N → ∞. I leave the detail to you as an exercise (Exercise 8).
2.12. Now we apply Cauchy’s criterion to convergence of series. Recall that a series
∞
n= 1
an is convergent if its partial sums sn ≡ a1 + a2 + · · · + an (n = 1, 2, . . .) form
a convergent sequence and, in this case, s ≡ limn→ ∞ sn is called the sum of the series
∞
n= 1 an . Cauchy’s criterion tells us that {sn } converges if and only if there is a sequence
of positive numbers αn decreasing to zero such that |sm −sn | < αn for all n, m with m > n.
The last inequality can be rewritten as |an+
1 +an+ 2 +· · · +am |
< αn . With a slight change
in notation, we have
Cauchy’s Criterion for Series. A series
∞
n= 1
an is convergent if and only if there
is a sequence of positive numbers αn decreasing to zero such that |an +an+
1 +· · ·+am |
< αn
for all m, n with m > n.
2.13. A consequence of Cauchy’s criterion for series is the following
pn with positive terms
Fact 1:
an is convergent if there is a convergent series
pn > 0, a number K > 0 and an integer N > 0 such that |an | ≤ Kpn for all n > N .
Proof: Since
pn is a convergent series with positive terms, there is a sequence of
positive numbers αn decreasing to zero such that pn + pn+
1
+ · · · + pm < αn for all n, m
with m > n. Thus, for n, m with m > n > N , we have
|an + an+ 1 + · · · + am | ≤ |an | + |an+ 1 | + · · · + |am | ≤ Kpn +Kpn+ 1 + · · · +Kpm < Kαn .
Notice that {Kαn }n is also a sequence decreasing to zero. Hence, by Cauchy’s criterion,
we know that the series ∞n= N+ 1 an converges. Hence the series ∞n= 1 an also converges.
8
The above fact may be called the comparison test. Usually it is applied in the
following way. Given a series
an we compare it with a positive series
pn which is
known to be convergent. If we can show that the terms of the given series can be controlled
by the terms of this positive series in the sense that |an | ≤ Kpn for all n where K is some
positive number independent of n, then we can conclude that the given series is convergent.
∞
∞
n
n
For example, the series
is convergent, since
n= 1 cos nθ/2
n= 1 1/2 < +∞ and
| cos nθ/2n | ≤ 1/2n for all n. A special case of Fact 1 is the following:
Fact 2. A series
∞
n= 1 an
is convergent if
∞
n= 1 |an |
< ∞.
∞
This follows from Fact 1 by taking pn = |an | and K = 1. A series n= 1 an is said to
be absolutely convergent if ∞n= 1 |an | < ∞. Thus, Fact 2 says that an absolutely
convergent series must be convergent. A series is said to be conditionally convergent if
it is convergent but not absolutely convergent. We will see some examples of conditional
convergent series in the future.
2.14. Next we give two famous tests for absolute convergence:
Fact 3.
an is absolutely convergent if one of the following conditions holds:
(a) limn
|an+1 |
|an |
< 1. (Ratio Test) (b) limn |an |1/n < 1. (Root Test)
Proof: Assume (a). Take any real number r such that limn |an+
there is some positive number N such that |an+
1 |/|an |
1 |/|an |
< r < 1. Then
≤ r for all n ≥ N . Now, for n > N ,
we have
|aN +
1 |/|aN |
≤ r, |aN+
2 |/|aN + 1 |
≤ r, . . . . . . , |an |/|an−1 | ≤ r.
to |an−1 |, we get |an |/|aN | ≤ rn−N ,
∞
n
or |an | ≤ Krn , where K = |aN |r−N . Since 0 < r < 1,
n= 1 r is a convergent geometric
Multiplying all these inequalities and canceling |aN+
1|
series. Thus the given series, which can be compared to this convergent geometric series,
is absolutely convergent.
Assume (b). Choose r > 0 such that lim|an |1/n < r < 1. Then there exists some
positive integer N such that |an |1/n < r or |an | < rn for all n > N . The given series can
be compared to the convergent geometric series ∞n= 1 rn and hence converges. Q.E.D.
Example. Discuss the convergence or divergence of each of the following series.
∞
n= 1
nn
,
3n n!
∞
n= 1
9
nn
.
2n n!
[Remark: As n → ∞, the way that 2n and 3n “blow up” is slower than that of nn and
n!. But the “giants” nn and n! are fighting against each other and, as a consequence, the
small guys 2n and 3n showing up here actually decide the “fates” of these two series.] Let
us use the ratio test. For the first series, the ratio of consecutive terms an and an+
n
n
an+ 1
(n + 1)n+ 1 /3n+ 1 (n + 1)!
1 n+1
1
1
=
=
=
1+
an
nn /3n n!
3
n
3
n
1
is
which tends to e/3 ≈ 2.2/3 < 1 as n → ∞. Hence the first series converges. Now, by
a similar computation, we find that the ratio an+
1 /an
of consecutive terms of the second
series tends to e/2, which is greater 1. This implies that the nth term nn /2n n! of this
series blows up as n → ∞. So, actually the second series is “extremely divergent”.
2.15. We present a useful theorem attributed to Abel. The proof of this theorem involves an interesting technique called “summation by parts”, which is the discrete analogue
of the well-known “integration by parts”.
THEOREM (Abel). If {an } is a sequence of positive numbers decreasing to zero,
∞
and if the partial sums of the series n bn are bounded, then n= 1 an bn converges.
Remark: Condition (b) of the theorem means that there is a positive number M > 0 such
that |b1 + b2 + · · · + bn | < M for all n. Notice that this condition is much weaker than the
convergence of the series n bn .
Proof: The Cauchy criterion suggests us to look at the sums
sn,m ≡ an bn + an+
1 bn+ 1
+ · · · + am bm ,
for large m and n with m > n. For each n, let Sn = b1 + b2 + · · · + bn . Then there is
M > 0 such that |Sn | ≤ M for all n. Notice that bn = Sn − Sn−1 for all n > 1. Now sn,m
can be rewritten as
an (Sn − Sn−1 ) + an+
1 (Sn+ 1 −
Sn ) + an+
= − an Sn−1 + an Sn − an+ 1 Sn + an+ 1 Sn+
1
2 (Sn+ 2 −
− an+ 2Sn+
= − an Sn−1 +(an −an+ 1 )Sn +(an+ 1 −an+ 2 )Sn+
1
1
Sn+
1)
+ · · · + am (Sm − Sm−1 )
+ · · · + am−1 Sm−1 − am Sm−1 + am Sm
+ · · · + (am−1 −am )Sm−1 + am Sm
Since {an } is a decreasing sequence of positive numbers, we can estimate |sn,m | as follows:
|sn,m | ≤ an |Sn−1 |+(an −an+ 1 )|Sn |+(an+ 1 −an+ 2 )|Sn+ 1 |+· · ·+(am−1 −am )|Sm−1 |+am |Sm |
≤ an M +(an −an+ 1 )M +(an+ 1 −an+ 2 )M +· · ·+(am−1 −am )M +am M
= 2M an .
10
Since 2M an decreases to zero,
n an bn
convergest according to Cauchy. Q.E.D.
2.16. An immediate consequence of the above theorem is the following:
Corollary (Alternating Series Test). If {an }n≥0 is a sequence of positive numbers
∞
decreasing to zero, then the series n= 0 (−1)n an converges.
To prove this corollary, just let bn = (−1)n in the above theorem and notice that the
partial sums of n (−1)n is bounded by 1.
Now it is easy to construct a conditional convergent series. Take any sequence {an }
which decreases to zero such that n an = +∞, e.g. the following sequence in which each
1
n
appears n times: 1, 12 , 12 , 13 , 13 , 13 , 14 , 14 , 14 , 14 , . . .. By the alternating series test, we know
∞
that the series n= 1 (−1)n an converges. But, by our choice of {an }, this series is not
absolutely convergent.
2.17. Example. Let {an } be a sequence of positive numbers decreasing to zero and
let θ be an arbitrary real number. Consider the series
an cos nθ. If
an < +∞, then,
∞
by the comparison test, we know that the series
n= 1 an cos nθ is absolutely convergent.
Hence in what follows we assume
an = +∞. Since cos nθ is an even function with
2π as a period, we may assume that 0 ≤ θ ≤ π. When θ = 0, the series
an cos nθ
becomes
n an , which diverges to +∞. So we assume that 0 < θ ≤ π. Notice that, under
this assumption, we have cos θ < 1. Now we apply Abel’s theorem above to show that the
series
an cos nθ converges. To this end, we need the boundedness of the partial sums
of the series n cos nθ. We proceed to find a closed form of these partial sums to carry
out our upper bound estimate. The most convenient way of finding this closed form is by
using complex numbers.
Let z = cos θ + i sin θ. By DeMoivre’s theorem, we have z n = cos nθ + i sin nθ. So
(1 + cos θ+ cos 2θ + · · · + cos nθ) + i(sin θ + sin 2θ + · · · + sin nθ)
1 − z n+ 1
.
1−z
(Notice that, to get the last expression, we need cos θ < 1 to ensure z = 1.) Hence
1 − z n+ 1 2
≤
|1 + cos θ + cos 2θ + · · · + cos nθ| ≤ 1−z
|1 − z|
∞
for all n. Thus the partial sums of
n= 0 cos nθ form a bounded sequence and Abel’s
theorem above applies. We conclude that the series
n an cos nθ converges if θ is any
= 1 + z + z2 + · · · + zn =
number not an integral multiple of 2π and diverges to +∞ otherwise.
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