ma1010-notes-1
DESCRIPTION
Thamban Nair notesTRANSCRIPT
Sequence and Series of Numbers∗
M.T.Nair (IIT Madras)
July 23, 2014
Contents
1 Sequence of numbers 2
1.1 Convergence and divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Some tests for convergence and divergence . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.3 Monotonic sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.4 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.5 Further examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.6 Cauchy criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2 Series of Real Numbers 25
2.1 Convergence and divergence of series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.2 Some tests for convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.3 Alternating series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.4 Absolute convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
∗Taken from the forthcoming book Calculus of One Variable by M.Thamban Nair
1
1 Sequence of numbers
Suppose for each positive integer n, we are given a real number an. Then, the list of numbers
a1, a2, . . . , an, . . .
is called a sequence. A more precise definition of a sequence is the following:
Definition 1.1. A sequence of real numbers is a function from the set N of natural numbers to the
set R of real numbers. ♦
If f : N→ R is a sequence, and if an = f(n) for n ∈ N, then we write the sequence f as
(a1, a2, . . .) or (an) or {an},
and the term an is called the nth term of the sequence (an). In specific cases one may also write a
sequence as
(a1, a2, . . . , an, . . .) instead of (a1, a2, . . .).
Remark 1.2. A sequence (a1, a2, . . .) is different from the set {an : n ∈ N}. For instance, a number
may be repeated in a sequence (an), but it need not be written repeatedly in the set {an : n ∈ N}. As
an example, (1, 12 , 1,13 , . . . , 1,
1n , . . .) is a sequence (an) with a2n−1 = 1 and a2n = 1/(n + 1) for each
n ∈ N, whereas the set {an : n ∈ N} is same as the set {1/n : n ∈ N}. ♦
Example 1.3. Let us consider a few examples of sequences:
(i) (an) with an = 1 for all n ∈ N.
(ii) (an) with an = n for all n ∈ N.
(iii) (an) with an = 1/n for all n ∈ N.
(iv) (an) with an = n/(n+ 1) for all n ∈ N.
(v) (an) with an = (−1)n for all n ∈ N – the sequence takes values 1 and −1 alternately. ♦
1.1 Convergence and divergence
In certain sequences the nth term comes closer and closer to a particular number as n becomes larger
and larger. For example, in the sequence ( 1n ), the nth term comes closer and closer to 0, whereas in
( nn+1 ), the nth term comes closer and closer to 1 as n becomes larger and larger. If you look at the
sequence ((−1)n), the terms oscillate between −1 and 1 as n varies, whereas in (n2) the terms become
larger and larger.
2
What do we mean by the statement “an comes closer and closer to a number a as n
becomes larger and larger”?
Saying differently: an can be made arbitrarily close to a by taking n large enough.
Definition 1.4. A sequence (an) of real numbers is said to converge to a real number a if for every
ε > 0, there exists a positive integer N (in general depending on ε) such that
|an − a| < ε ∀ n ≥ N.
A sequence which converges is called a convergent sequence, and a sequence which does not converge
is called a divergent sequence. ♦
Suppose (an) converges to a. Then as per the above definition, taking ε = 10−k for some k ∈ N,
there exists some Nk ∈ N such that
|an − a| < 10−k ∀ n ≥ Nk.
In other words:
If n ≥ Nk and if an and a are expressed in decimal expansion, then
the first k decimal places of an and a are the same.
• If (an) converges to a, then we write
an → a as n→∞
or, in short,
an → a,
which we may read as “an tends to a as n tends to infinity”.
• If (an) does not converge to a, then we write an 6→ a.
It is easily seen that (verify)
an → a ⇐⇒ |an − a| → 0 ⇐⇒ |an| → |a|.
Remark 1.5. We must keep in mind that the symbol ∞ is not a number; it is only a notation used
in the context of describing some properties of real numbers, such as in Definition 1.4. ♦
Note that, in the definition of convergence, different ε can result in different N , i.e., the number
N may vary as ε varies. This fact is illustrated in the following examples.
3
Example 1.6. Consider the sequence (an) with an = 1/n for n ∈ N. We expect that an → 0 as
n→∞. Let us prove this: So, let ε > 0 be given. Then
1
n< ε ⇐⇒ n >
1
ε.
Thus, taking a positive integer N > 1/ε, we obtain |an − 0| < ε for all n ≥ N . Similarly, for any
ε > 0, ∣∣∣ (−1)n
n− 0∣∣∣ =
1
n< ε ∀n ≥ N,
where N is a positive integer such that N > 1/ε. Thus we have shown that
1
n→ 0 and
(−1)n
n→ 0 as n→∞.
Note that the positive integer N in both the examples depend on the ε chosen. ♦
Example 1.7. Consider the sequence (an) with an = n/(n+ 1) for n ∈ N. We claim that an → 1 as
n→∞. Note that
|an − 1| = 1
n+ 1so that, for ε > 0,
|an − 1| < ε ⇐⇒ n+ 1 >1
ε.
Thus, taking a positive integer N with N ≥ 1/ε, we obtain |an − 1| < ε for all n ≥ N . Note that,
n ≥ 100 =⇒ |an − 1| < 1
100,
n ≥ 1000 =⇒ |an − 1| < 1
1000.
More generally, for a given k ∈ N,
|an − 1| < 1
10k∀n ≥ 10k.
It is also clear that |an − 1| ≥ 1/10k if n < 10k. ♦
Example 1.8. For n ∈ N, let
an =9
10+
9
102+ · · ·+ 9
10n,
i.e., an = 0.99 · · · 9 with 9 repeated n times. You may guess that an → 1 as n → ∞. This is true:
Note that
9
10+
9
102+ · · ·+ 9
10n=
9
10
(1 +
1
10+ · · ·+ 1
10n−1
)=
9
10
[1− 1
10n
1− 110
]= 1− 1
10n.
Hence, |an − 1| = 1/10n so that
|an − 1| < ε ⇐⇒ 10n >1
ε.
Let N ∈ N be such that 10N > 1/ε. Then, we have |an − 1| < ε for all n ≥ N . Therefore, an → 1 as
n→∞. ♦
4
Example 1.9. For n ∈ N, let
an =3
10+
3
102+ · · ·+ 3
10n,
i.e., an = 0.33 · · · 3 with 3 repeated n times. Note that
3
10+
3
102+ · · ·+ 3
10n=
3
10
[1− 1
10n
1− 110
]=
1
3
[1− 1
10n
].
Thus, |an − 1/3| = 1/(3 × 10n) so that, for any given ε > 0, we can choose a positive integer N
(depending on ε) such |an − 1/3| < ε for all n ≥ N . Therefore, an → 1/3 as n→∞. ♦
In the examples above we have used an important property of real numbers, known as the
Archimedean Property :
Archimedean Property : Given any x ∈ R, there exists n ∈ N such that n > x.
We shall use this property throughout the text without mentioning it explicitly.
Let us show the convergence of a few more sequences which we encounter very often.
Exercise 1.10. Using the definition of convergence, show that1
n2→ 0 and
1
2n→ 0. ♦
It can happen that for a convergent sequence (an), same N can work for different ε’s. To
illustrate this point, look at the following examples.
Example 1.11. Let an = 1 for all n ∈ N. We can easily guess that (an) converges to 1. Note that
for any ε > 0,
|an − 1| < ε ∀n ≥ 1.
Thus, N = 1 works for any ε > 0. ♦
Example 1.12. Let an = n for all n ∈ {1, 2, . . . , 99} and an = 1 for all n ≥ 100. You must have
guessed that, (an) converges to 1. Yes; it is true: For any given ε > 0,
|an − 1| < ε ∀n ≥ 100.
Thus, N = 100 works for any ε > 0. ♦
Definition 1.13. (i) A sequence is said to be a constant sequence if all its terms are the same,
i.e., an = a1 for all n ∈ N.
(ii) A sequence is said to be an eventually constant sequence if all its terms after a particular
stage are the same, i.e., there exists k ∈ N such that an = ak for all n ≥ k. ♦
Remark 1.14. We may observe that, for ε > 0,
|an − a| < ε ⇐⇒ an ∈ (a− ε, a+ ε).
5
Thus, (an) converges to a ∈ R if and only if for every ε > 0, there exists a positive integer N such
that
an ∈ (a− ε, a+ ε) ∀n ≥ N.
In other words, an → a as n → ∞ if and only if for every ε > 0, an belongs to the open interval
(a− ε, a+ ε) for all n after some finite stage, and this finite stage may vary according as ε varies. ♦
The above remark leads to the following (verify):
1. an → a if and only if for every open interval I containing a, there exists a positive integer N
(depending on I) such that an ∈ I for all n ≥ N .
2. an 6→ a if and only if there is an open interval I containing a such that infinitely may an’s are
not in I.
Now, let us observe an important property of convergent sequences.
THEOREM 1.15. If (an) converges to a, then (an) cannot converge to b with b 6= a.
Proof. Suppose an → a and an → b as n → ∞, and suppose that b 6= a. Let I1 and I2 be disjoint
open intervals such that a ∈ I1 and b ∈ I2. Then, there exist positive integers N1 and N2 such that
an ∈ I1 ∀n ≥ N1 and an ∈ I2 ∀n ≥ N2.
Thus, an ∈ I1 ∩ I2 for all n ≥ N := max{N1, N2}; a contradiction, since I1 ∩ I2 = ∅.
Thus, we can define the concept of the limit of a convergent sequence.
Definition 1.16. If (an) converges to a, then a is called the limit of (an), and we write
limn→∞
an = a.
♦
Exercise 1.17. an → a if and only if ak+n → a as n→∞ for any k ∈ N. ♦
Now, let us give a few examples of divergent sequences.
Example 1.18. Let an = (−1)n+1 for n ∈ N. The terms of the sequence change alternately between
1 and −1. So, one may guess that it cannot converge to any number. To see this, we observe that, for
any a ∈ R, if we take 0 < ε < 1/2, either 1 or −1 will be outside the interval (a − ε, a + ε); in other
words, infinitely many of the terms of the sequence (an) lie outside (a− ε, a+ ε). ♦
Example 1.19. Let (an) be defined by a2n−1 = 1/n and a2n = 1. If we take ε a positive number less
than 1, then, for any a ∈ R, infinitely many of the terms of the sequence (an) lie outside (a− ε, a+ ε).
For instance, let us take ε = 1/2. If a ≥ 1/2, then infinitely many of a2n−1 lie outside (a− ε, a+ ε),
and if a < 1/2, then every a2n lie outside (a− ε, a+ ε) for all n ∈ N. Thus, an 6→ a for any a ∈ R. ♦
6
Example 1.20. Let an = n for n ∈ N. In this case, for any given ε > 0 and for any a ∈ R, infinitely
many of the terms of the sequence (an) lie outside (a − ε, a + ε). Similar is the case if an = n2 or
an = n2/(n+ 1) (verify). Thus, an 6→ a for any a ∈ R. ♦
Definition 1.21. A sequence (an) is said to be an alternating sequence if an changes sign alter-
nately, that is, anan+1 < 0 for every n ∈ N. ♦
An alternating sequence may converge or diverge. We have seen that the sequence ((−1)n) diverges,
whereas ((−1)n/n) converges to 0.
A divergent sequence may have some specific properties such as the ones defined below.
Definition 1.22. Consider a sequence (an).
(1) We say that (an) diverges to infinity if for every M > 0, there exists N ∈ N such that
an > M ∀ n ≥ N,
and write this fact as an →∞.
(2) We say that (an) diverges to minus infinity if for every M > 0, there exists N ∈ N such that
an < −M ∀n ≥ N
and write this fact as an → −∞.
♦
Example 1.23. (i) The sequence (an) with an = n2/(n+ 1) diverges to infinity: Note that
n2
n+ 1≥ n
2∀n ∈ N.
(ii) The sequence (an) with an = 2n diverges to infinity: Note that 2n ≥ n for all n ∈ N.
Observe the following facts:
1. If an →∞ or an → −∞ as n→∞, then (an) is a divergent sequence.
2. an →∞ if and only if −an → −∞.
3. If (an) and (bn) are such that an ≤ bn for all n ∈ N, then
(a) an →∞ as n→∞ implies bn →∞ as n→∞,
(b) bn → −∞ as n→∞ implies an → −∞ as n→∞.
7
Exercise 1.24. Prove the above statements. ♦
The following theorem is useful for showing convergence of certain sequences using the convergence
of some other sequences.
THEOREM 1.25. Suppose an → a and bn → b as n→∞. Then the following results hold.
(i) an + bn → a+ b as n→∞,
(ii) c an → c a as n→∞ for any real number c,
(iii) If an ≤ bn for all n ∈ N, then a ≤ b.
(iv) (Sandwich theorem) If an ≤ cn ≤ bn for all n ∈ N, and if a = b, then cn → a as n→∞.
Proof. Let ε > 0 be given.
(i) Note that, for every n ∈ N,
|(an + bn)− (a+ b)| = |(an − a) + (bn − b)|
≤ |an − a|+ |bn − b|.
Since an → a and bn → b, the above inequality suggests that we may take ε1 = ε/2, and consider
N1, N2 ∈ N such that
|an − a| < ε1 ∀n ≥ N1 and |bn − b| < ε1 ∀n ≥ N2
so that
|(an + bn)− (a+ b)| ≤ |an − a|+ |bn − b| < 2ε1 = ε
for all n ≥ N := max{N1, N2}.
(ii) Note that
|can − ca| = |c| |an − a| ∀n ∈ N.
If c = 0, then (can) is a constant sequence with every term 0, and hence can → 0. Next, suppose
c 6= 0. Then we have
|can − ca| < ε ⇐⇒ |an − a| <ε
|c|.
Since an → a, there exists N ∈ N such that
|an − a| <ε
|c|∀n ≥ N.
Thus,
|can − ca| < ε ∀n ≥ N.
8
(iii) Suppose an ≤ bn for all n ∈ N. Let N1, N2 ∈ N be such that
|an − a| < ε ∀n ≥ N1 and |bn − b| < ε ∀n ≥ N2.
Hence,
a− ε < an < a+ ε and b− ε < bn < b+ ε ∀n ≥ N = max{N1, N2}.
In particular,
a− ε < an ≤ bn < b+ ε ∀n ≥ N.
Thus, we obtain a < b+ 2ε. Since this is true for every ε > 0, we have a ≤ b.
(iv) Suppose an ≤ cn ≤ bn for all n ∈ N and a = b. We have to show that cn → a. Note that
cn − an ≤ bn − an = (bn − a) + (a− an) ∀n ∈ N.
Hence,
0 ≤ |cn − an| ≤ |bn − a|+ |a− an| ∀n ∈ N.
Therefore, from the fact that an → a and bn → a and the result in (iii), we obtain |cn − an| → 0 so
that by (i), cn = an + (cn − an)→ a.
Suppose an → a and bn → b as n→∞. We shall prove:
1. anbn → ab and
2. an/bn → a/b whenever bn 6= 0 for all n ∈ N and b 6= 0.
The following theorem together with Theorem 1.25 can be used for inferring convergence or diver-
gence of certain sequences.
THEOREM 1.26. The following results hold.
(i) If a > 1, then an →∞.
(ii) If 0 < a < 1, then an → 0.
Proof. (i) Suppose a > 1. Writing a = 1 + r with r > 0, we have
an = (1 + r)n = 1 + nr +n(n− 1)
2!r2 + · · ·+ rn
for all n ∈ N. Hence, an ≥ rn for all n ∈ N. Since rn→∞, we have an →∞.
(ii) Let 0 < a < 1 and ε > 0 be given. Then b = 1/a satisfies b > 1 so that bn →∞. Hence there
exists N ∈ N such that bn > 1/ε for all n ≥ N . But, bn > 1/ε if and only if an < ε. Hence,
0 < an < ε ∀n ≥ N.
Thus, an → 0.
9
Example 1.27. For 0 < r < 1 and n ∈ N, let an = 1 + r + r2 + · · ·+ rn. Then
an =1− rn+1
1− r→ 1
1− ras n→∞.
Exercise 1.28. Show that |a| < 1 implies an → 0. ♦
Exercise 1.29. For a ≥ 0, prove the following.
(i) an → 0 ⇐⇒ a < 1.
(ii) an →∞ ⇐⇒ a > 1. ♦
Exercise 1.30. If an → a and a 6= 0, then show that there exists k ∈ N such that an 6= 0 for all
n ≥ k. ♦
Exercise 1.31. Show that
an :=(
1 +1
n
)1/n→ 1.
[Hint: Observe that 1 ≤ an ≤ (1 + 1/n) for all n ∈ N.] ♦
1.2 Some tests for convergence and divergence
THEOREM 1.32. (Ratio test) Suppose an > 0 for all n ∈ N such that limn→∞
an+1
an= ` for some
` ≥ 0. Then the following hold.
(i) If ` < 1, then an → 0.
(ii) If ` > 1, then an →∞.
Proof. (i) Suppose ` < 1. Let q be such that ` < q < 1. Then there exists N ∈ N be such that
an+1
an≤ q ∀n ≥ N.
Hence,
0 ≤ an ≤ qn−NaN ∀ n ≥ N.
By Theorem 1.26, qn−N → 0 as n→∞. Hence, by Sandwich theorem, an → 0 as n→∞.
(ii) Suppose ` > 1. Let q be such that 1 < q < `. Then there exists N ∈ N be such that
an+1
an> q ∀n ≥ N.
Hence,
an ≥ qn−NaN ∀ n ≥ N.
By Theorem 1.26, qn−N →∞ as n→∞. Hence, an →∞.
10
Example 1.33. Let 0 < a < 1. Then nan → 0 as n → ∞. To see this, let an := nan for n ∈ N.
Then we havean+1
an=
(n+ 1)an+1
nan=
(n+ 1)a
n∀ n ∈ N.
Hence, limn→∞
an+1
an= a < 1. Thus, by Theorem 1.32, nan → 0.
Similarly, it can be shown (give details) that, for any k ∈ N, nkan → 0 as n→∞. ♦
Remark 1.34. The converse of the results (i) and (ii) in Theorem 1.32 does not hold. To see this
first consider the following examples:
(i) Consider (an) with an = 1/n. Then an → 0, but an+1/an → 1.
(ii) Consider (an) with an = n. Then an →∞, but an+1/an → 1.
In fact, we shall see in the next section that the condition ` < 1 in Theorem 1.32 is too strong, in
the sense that, not only we have the convergence of (an) to 0, but also we can show the convergence
of the sequence (sn), where sn = a1 + a2 + · · · + an. Note that if sn → s for some s ∈ R, then
an = sn − sn−1 → 0. ♦
The following theorem gives a sufficient condition for certain number to be a limit of a given
sequence.
THEOREM 1.35. Let (an) be a sequence such that
|an+1 − a| ≤ r|an − a| ∀n ∈ N
for some a ∈ R and for some r with 0 < r < 1. Then an → a.
Proof. For each n ∈ N, we have
|an+1 − a| ≤ r|an − a| ≤ · · · ≤ rn|a1 − a|.
Since 0 < r < 1, by Theorem 1.26, rn → 0 so that by Sandwich theorem, an → a.
Example 1.36. Let a sequence (an) be defined iteratively as follows :
a1 = 1, an+1 =2 + an1 + an
, n = 1, 2, . . . .
Let us assume for a moment that (an) converges to a ∈ R. Then we obtain
a =2 + a
1 + a,
i.e., a(1 + a) = 2 + a, i.e., a2 = 2. Thus, if (an) converges, then the limit must be a =√
2. Now, we
prove that (an) actually converges to a :=√
2. Note that
an+1 − a =2 + an1 + an
− 2 + a
1 + a
=(2 + an)(1 + a)− (1 + an)(2 + a)
(1 + an)(1 + a)
=a− an
(1 + an)(1 + a).
11
Note that an ≥ 1 for all n ∈ N and a ≥ 1. Hence, we obtain
|an+1 − a| ≤|a− an|
(1 + an)(1 + a)≤ |an − a|
4∀n ∈ N.
Therefore, by Theorem 1.35, an → a =√
2. ♦
Remark 1.37. In view of Theorem 1.35, one may ask the following question:
If (an) is such that |an+1 − a| < |an − a| for all n ∈ N for some a ∈ R, then does (an)
converge to a?
Not necessarily! To see this, consider the sequence (an) with
an =n+ 1
n, n ∈ N.
Sincen+ 2
n+ 1<n+ 1
n∀n ∈ N,
taking a = 0, we have |an+1 − a| < |an − a| for all n ∈ N. But (an) does not converge to 0. In fact,
an → 1.
One may also enquire if the condition on (an) in Theorem 1.35 can be replaced by
|an+2 − an+1| ≤ r|an+1 − an| ∀n ∈ N
for some r with 0 < r < 1. In this case, the answer is in the affirmative. We shall present its proof at
the end of the present section after introducing the concept of a Cauchy sequence. ♦
Exercise 1.38. Let an+1 = 1 +1
anfor all n ∈ N. Prove the following:
(i) an → a ⇐⇒ a =1 +√
5
2.
(ii) a =1 +√
5
2satisfies a = 1 + 1/a.
(iii) an+1 − a| ≤ r|an − a| for some r with 0 < r < 1. ♦
Remark 1.39. The terms of the sequence (an) in the above exercise
1
1,
2
1,
3
2,
5
3,
8
5,
13
8, . . . .
These terms are ratios of consecutive terms of the
1, 1, 2, 3, 5, 8, 13, . . .
which is the well-known Fibonacci sequence1. The number
1 +√
5
2,
1Named after the Italian mathematician Leonardo Fibonacci (1170 – 1250).
12
which is the limit of the sequence (an), is called the golden ratio2. The golden ratio is approximately
equal to 1.618033 which is correct to (1 +√
5)/2 up to six decimal places, i.e.,∣∣∣1 +√
5
2− 1.618033
∣∣∣ < 1
106.
In fact, the number (1 +√
5)/2 is the ratio of the sides α and β with β < α of a rectangle such that
when the square of side β is cut out of the rectangle, the remaining rectangle has the same ratio as
its sides, i.e,α
β=
β
α− β.
Indeed, if γ = α/β, then the above relation gives γ = 1/(γ−1), i.e., γ = 1+1/γ so that γ = (1+√
5)/2.
Such a rectangle is called a golden rectangle. ♦
Next we prove a necessary condition for the convergence of a sequence.
THEOREM 1.40. (Boundedness test) If (an) converges, then there exists M > 0 such that
|an| ≤M ∀n ∈ N.
Proof. Suppose an → a. Note that, for all n ∈ N,
|an| = |(an − a) + a| ≤ |an − a|+ |a|.
Since |an − a| → 0, there exists N ∈ N such that |an − a| ≤ 1 for all n ≥ N . Then we have
|an| ≤ 1 + |a| ∀n ≥ N.
Thus, taking M = max{1 + |a|, |a1|, |a2|, . . . , |aN |}, we have |an| ≤M for all n ∈ N.
The converse of Theorem 1.40 is not true. Note that the sequence (an) with an = (−1)n satisfies
|an| = 1 for all n ∈ N, but it is not convergent.
For later use we introduce the following definition.
Definition 1.41. A sequence (an) is said to be
(1) bounded above if there exists a real number M such that an ≤M for all n ∈ N;
(2) bounded below if there exists a real number M ′ such that an ≥M ′ for all n ∈ N.
A sequence which is bounded above and bounded below is said to be a bounded sequence.
A sequence which is not bounded is called an unbounded sequence. ♦2It is known from antiquity that constructions based on approximate values of golden ratio are aesthetically pleasing.
13
We observe the following:
1. A sequence (an) is bounded if and only if there exists M > 0 such that
|an| ≤M ∀n ∈ N.
2. A sequence (an) is unbounded if and only if it is either not bounded above or not bounded
below.
3. If (an) diverges to ∞, then it is not bounded above, and if it diverges to −∞, then it is not
bounded below.
Thus, according to Theorem 1.40:
Every convergent sequence is bounded.
Example 1.42. Let us look at a few examples of bounded and unbounded sequences (verify).
(i) (an) with an = (−1)n is a bounded sequence.
(ii) (an) with an = (−1)nn is neither bounded above nor bounded below, and it neither diverges
to +∞ nor diverges to −∞.
(iii) (an) with an = −n is bounded above, but not bounded below, and it diverges to −∞.
(iv) (an) with an = n is bounded below, but not bounded above, and it diverges to +∞. ♦
Theorem 1.40 can be used to show that certain sequence is not convergent, as in the following
example.
Example 1.43. For n ∈ N, let
an = 1 +1
2+
1
3+ . . .+
1
n.
Then (an) diverges: To see this, observe that
a2n = 1 +1
2+
1
3+ . . .+
1
2n
= 1 +1
2+(1
3+
1
4
)+(1
5+
1
6+
1
7+
1
8
)+
. . .+( 1
2n−1 + 1+ . . .+
1
2n
)≥ 1 +
n
2.
Hence, (an) is not a bounded sequence, so that it diverges. ♦
14
Theorem 1.40 will be used for proving the following.
THEOREM 1.44. Suppose an → a and bn → b. Then we have the following.
(i) anbn → ab as n→∞.
(ii) If bn 6= 0 for all n ∈ N and b 6= 0, then
1
bn→ 1
band
anbn→ a
b.
Proof. (i) Note that, for every n ∈ N,
|anbn − ab| = |an(bn − b) + (an − a)b|
≤ |an| |bn − b|+ |an − a| |b|.
By Theorem 1.40, (an) is bounded, say |an| ≤M for all n ∈ N. Hence, we have
0 ≤ |anbn − ab| ≤M |bn − b|+ |b||an − a|.
Since an → a and bn → b, by Theorem 1.25, we obtain anbn → ab as n→∞.
(ii) Suppose bn 6= 0 for all n ∈ N and b 6= 0. Note that, for every n ∈ N,∣∣∣∣ 1
bn− 1
b
∣∣∣∣ =|bn − b||bn| |b|
and
|bn| = |b+ (bn − b)| ≥ |b| − |bn − b|
Since bn → b and b 6= 0, there exists N ∈ N be such that
|bn − b| < |b|/2 ∀n ≥ N.
Then,
|bn| = |b+ (bn − b)| ≥ |b| − |b|/2 = |b|/2 ∀n ≥ N.
Thus, we obtain ∣∣∣∣ 1
bn− 1
b
∣∣∣∣ ≤ |bn − b|(|b|/2) |b|
=2
|b|2|bn − b| ∀n ≥ N.
Hence, by Theorem 1.25, we obtain 1/bn → 1/b as n→∞. Now, using (i), we also obtain an/bn → a/b
as n→∞.
Exercise 1.45. Prove the following.
(i) If (an) is not bounded above, then there exists a sequence (kn) of natural numbers such that
kn ≤ kn+1 for all n ∈ N and akn → +∞ as n→∞.
(ii) If (an) is not bounded below, then there exists sequence (kn) of natural numbers such that
kn ≤ kn+1 for all n ∈ N and akn → −∞ as n→∞. ♦
15
Exercise 1.46. Let (an) and (bn) are such that an ≤ bn for all n ∈ N. Prove the following:
(i) If (bn) is bounded above, then (an) bounded above.
(ii) If (an) is bounded below, then (bn) bounded below. ♦
1.3 Monotonic sequences
We have seen that a bounded sequence need not converge. However, we shall show that, if the terms
of a bounded sequence either increases or decreases, then the sequence does converge.
Definition 1.47. A sequence (an) is said to be a
(1) monotonically increasing sequence if an ≤ an+1 for all n ∈ N;
(2) monotonically decreasing sequence if an ≥ an+1 for all n ∈ N;
(3) monotonic sequence if it is either monotonically increasing or monotonically decreasing.
If strict inequality occur in (1) (resp. (2)), then we say that the sequence is strictly increasing
(resp. strictly decreasing).
A monotonically increasing (respectively, a monotonically decreasing) sequence is also called an
increasing (respectively, a decreasing) sequence. ♦
We may observe that:
1. A sequence (an) is monotonically increasing and bounded above if and only if (−an) is mono-
tonically decreasing and bounded below.
2. Every monotonically increasing sequence is bounded below, and every monotonically decreasing
sequence is bounded above.
Example 1.48. The following statements can be easily verified (verify):
(i) (an) with an = n/(n+ 1) is monotonically increasing.
(ii) (an) with an = (n+ 1)/n is monotonically decreasing.
(iii) (an) with an = (−1)nn/(n+1) is neither monotonically increasing nor monotonically decreas-
ing. ♦
Exercise 1.49. Let (an) be a monotonic sequence. Prove the following.
16
(i) Suppose (an) converges. Then it is bounded above by the limit if it is increasing, and bounded
below by the limit if it is decreasing.
(ii) Suppose (an) diverges. Then it diverges to either infinity or minus infinity depending on
whether it is increasing or decreasing. ♦
A convergent sequence need not be monotonically increasing or monotonically decreasing.
However, we have the following theorem.
THEOREM 1.50. (Monotone convergence theorem) Every bounded monotonic sequence is
convergent.
Exercise 1.51. (i) A monotonically increasing sequence either converges or diverges to ∞.
(ii) A monotonically decreasing sequence either converges or diverges to −∞. ♦
In the following examples, Theorem 1.50 has been used for showing convergence of certain se-
quences.
Example 1.52. We have already seen that if 0 < a < 1, then an → 0 as n→∞ (See Theorem 1.26).
This can also be seen by making use of Theorem 1.50, as follows: Let xn = an. Then
0 ≤ xn+1 = an+1 = axn ≤ xn ∀n ∈ N.
Thus, (xn) is monotonically decreasing and bounded below. Hence (xn) converges to some x ∈ R.
Then xn+1 = axn → ax. Therefore, x = ax so that x = 0. ♦
Example 1.53. For n ∈ N, let
an = 1 +1
22+
1
32+ · · ·+ 1
n2.
Clearly, (an) is monotonically increasing. Now, we show that it is bounded above, so that by Theorem
1.50, it is convergent: We note that1
22+
1
32≤ 2
22=
1
2,
1
42+
1
52+
1
62+
1
72≤ 4
42=
1
4,
and more generally,
1
(2n−1)2+
1
(2n−1 + 1)2+ · · ·+ 1
(2n − 1)2≤ 2n−1
(2n−1)2=
1
2n−1.
Hence,
a2n = 1 +1
22+
1
32+ · · ·+ 1
(2n)2
≤ 1 +1
2+
1
22+ · · ·+ 1
2n−1+
1
22n
= 2
[1− 1
2n
]+
1
22n
≤ 2.
17
Since an ≤ a2n for all n ∈ N, we obtain that (an) is monotonically increasing and also bounded above,
and hence, by Theorem 1.50, it converges. We shall see in the final chapter of this book that the
number to which this (an) converges is π2/6. ♦
Example 1.54. Given a sequence (an) with an ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, consider the sequence (bn)
with
bn =a110
+a2102
+ · · ·+ an10n
for n ∈ N. Does (bn) converge? By our knowledge in decimal expansion of a number, we can believe
that (bn) would converge to a number in the interval [0, 1]. Let us prove this:
Note that bn ≤ bn+1 for all n ∈ N. Therefore, to assert the convergence of (bn), by Theorem 1.50,
it is enough to show that (bn) is bounded above. For this end, we observe that
bn =a110
+a2102
+ · · ·+ an10n≤ 9
10+
9
102+ · · ·+ 9
10n=
9
10
[1− 1
10n
1− 110
]= 1− 1
10n.
Thus, (bn) is bounded above by 1. Therefore, (bn) converges to a number in [0, 1]. ♦
Example 1.55. (The number e) Consider the sequences (an) and (bn) defined by
an =
(1 +
1
n
)n, bn = 1 +
1
1!+
1
2!+
1
3!+ · · ·+ 1
n!.
We show that both (an) and (bn) are monotonically increasing and bounded above. Hence, by Theorem
1.50, they converge. Further, we shall show that they have the same limit. For this we show that
(i) bn ≤ bn+1 ≤ 3 for all n ∈ N,
(ii) an ≤ bn for all n ∈ N,
(iii) an ≤ an+1 ≤ 3 for all n ∈ N.
Clearly, bn ≤ bn+1 for all n ∈ N. Also,
bn ≤ 1 + 1 +1
2+
1
22+ · · ·+ 1
2n−1< 3.
Thus, (i) is proved. Next,
an = 1 + 1 +1
2!
n(n− 1)
n2+ · · ·+ 1
n!
n(n− 1) . . . 2.1
nn
≤ 1 + 1 +1
2!+
1
3!+ · · ·+ 1
n!= bn
18
and
an =(
1 +1
n
)n= 1 + n.
1
n+n(n− 1)
2!
1
n2+ · · ·+ n(n− 1) . . . 2.1
n!
1
nn
= 1 + 1 +1
2!
(1− 1
n
)+
1
3!
(1− 1
n
)(1− 2
n
)+
· · ·+ 1
n!
(1− 1
n
)· · ·(
1− n− 1
n
)≤ an+1.
Thus, the proofs of (i)-(iii) are over. From, (i)-(iii), we see that both (an) and (bn) are monotonically
increasing bounded above. Hence, by Theorem 1.50, both (an) and (bn) converge. Let a and b their
limits. We show that a = b.
We have already observed that an ≤ bn. Hence, taking limits, we obtain a ≤ b. Notice that
an = 1 + 1 +1
2!
(1− 1
n
)+
1
3!
(1− 1
n
)(1− 2
n
)+
. . .+1
n!
(1− 1
n
). . .(
1− n− 1
n
).
Hence, for m,n with m ≤ n, we have
an ≥ 1 + 1 +1
2!
(1− 1
n
)+
1
3!
(1− 1
n
)(1− 2
n
)+
· · ·+ 1
m!
(1− 1
n
). . .(
1− m− 1
n
).
Taking limit as n→∞, we get (cf. Theorem 1.25 (c))
a ≥ 1 +1
1!+
1
2!+
1
3!+ · · ·+ 1
m!= bm.
Now, taking limit as m→∞, we get a ≥ b. Thus we have proved a = b.
The common limit of the two sequence (an) and (bn) above is denoted by the letter e, after the
great mathematician Euler3. ♦
Further examples are given in Subsection 1.5.
1.4 Subsequences
Definition 1.56. A subsequence of a sequence (an) is a sequence of the form (akn), where (kn) is
a strictly increasing sequence of positive integers. ♦3Leonhard Euler (15 April 1707 – 18 September 1783) was a Swiss mathematician and physicist. He made important
discoveries in various fields in mathematics. He introduced many modern mathematical terminology and notation,
including the notion of a mathematical function (Curtsey to Wikipedia).
19
A sequence (bn) is a subsequence of a sequence (an) if and only if there is a strictly
increasing sequence (kn) of positive integers such that bn = akn for all n ∈ N.
For example, given a sequence (an), the sequences
(a2n), (a2n+1), (an2), (a2n)
are some of the subsequences of (an). As concrete examples,
(1) ( 12n ), ( 1
2n+1 ), ( 1n2 ) and ( 1
2n ) are subsequences of ( 1n );
(2) ( 1n ) and ( n
n+1 ) are subsequences of (1, 12 ,12 ,
23 ,
13 ,
34 , . . .).
(3) (2n), (2n+ 1), (n2) and (2n) are subsequences of the sequence (an) with an = n for all n ∈ N.
THEOREM 1.57. If a sequence (an) converges to a, then all its subsequences converge to the same
limit a.
Proof. Suppose an → a. Consider a subsequence (akn) of (an). Let ε > 0 be given. Since an → a,
there exists N ∈ N such that
|an − a| < ε ∀n ≥ N.
In particular, since kN ≥ N ,
|an − a| < ε ∀n ∈ {kN , kN+1, . . .}.
Thus,
|akn − a| < ε ∀n ≥ N.
Hence, akn → a.
What about the converse of the above theorem? Obviously, if all subsequences of (an) converge,
then (an) also has to converge, since (an) is a subsequence of itself. Thus, we have proved:
(an) converges to a iff every subsequence of (an) converges to a.
We know that a divergent sequence can have convergent subsequences. But, this cannot happen
for monotonic sequences.
If (an) is a monotonic sequence having at least one convergent
subsequence, say with limit a, then (an) itself converge to a.
20
Exercise 1.58. Prove the above statement. ♦
We have seen in Theorem 1.40 that every convergent sequence is bounded, but a bounded se-
quence need not be convergent. However, we have the following theorem, which is called the Bolzano-
Weierstrass theorem.
THEOREM 1.59. (Bolzano-Weierstrass theorem4). Every bounded sequence of real numbers
has at least one convergent subsequence.
1.5 Further examples
Example 1.60. Let a sequence (an) be defined iteratively as follows: a1 = 1 and
an+1 =2an + 3
4
n ∈ N. We show that (an) is monotonically increasing and bounded above.
Note that
an+1 =2an + 3
4=an2
+3
4≥ an ⇐⇒ an ≤
3
2.
Thus it is enough to show that an ≤ 3/2 for all n ∈ N.
Clearly, a1 ≤ 3/2. If an ≤ 3/2, then an+1 = an/2 + 3/4 < 3/4 + 3/4 = 3/2. Thus, we have proved
that an ≤ 3/2 for all n ∈ N. Hence, by Theorem 1.50, (an) converges. Let its limit be a. Then taking
limit on both sides of an+1 = 2an+34 we have
a =2a+ 3
4i.e., 4a = 2a+ 3 so that a =
3
2.
Another solution: Since a := 3/2 satisfies
a =2a+ 3
4,
we obtain
an+1 − a =2an + 3
4− 2a+ 3
4=
1
2(an − a).
Thus, by Theorem 1.35, an → a = 3/2. ♦
Example 1.61. Given an arbitrary positive real number a0, let a sequence (an) be defined as follows
: a1 = 2 and
an+1 =1
2
(an +
2
an
)for n ∈ N. Note that, if the sequence converges, then its limit a ≥ 0, and then
a =1
2
(a+
2
a
)4Bernard Bolzano (October 5, 1781 - December 18, 1848), was a Bohemian mathematician, logician, philosopher,
theologian and Catholic priest of Italian Origins, and Karl Weierstrass (October 31, 1815 February 19, 1897) was a
German mathematician who is often cited as the “father of modern analysis”.
21
so that a =√
2.
Since a1 = 2, one may try to show that (an) is monotonically decreasing and bounded below.
Note that
an+1 :=1
2
(an +
2
an
)≤ an ⇐⇒ a2n ≥ 2, i.e., an ≥
√2,
and
an+1 :=1
2
(an +
2
an
)≥√
2 ⇐⇒ a2n − 2√
2an + 2 ≥ 0
⇐⇒ (an −√
2)2 ≥ 0.
Hence, an+1 ≥√
2 for all n ∈ N so that an+1 ≤ an for all n ≥ 2. Hence, (an) is monotonically
decreasing and bounded below, so that by Theorem 1.50, (an) converges and hence its limit is√
2. ♦
Example 1.62. The sequence (n1/n) converges and the limit is 1:
For each n ∈ N, since n1/n ≥ 1, there exists rn ≥ 0 such that
n1/n = 1 + rn.
Then we have
n = (1 + rn)n ≥ n(n− 1)
2r2n,
so that
r2n ≤2
n− 1∀n ≥ 2.
Since 2/(n− 1)→ 0, by Theorem 1.25(c), rn → 0. Hence n1/n → 1. ♦
Example 1.63. For any a > 0, (a1/n) converges to 1:
If a > 1, then we can write a1/n = 1 + rn for some sequence (rn) of positive reals. Then we have
a = (1 + rn)n ≥ nrn so that rn ≤ a/n.
Since a/n→ 0, by Theorem 1.25(c), rn → 0 and a1/n = 1 + rn → 1.
In case 0 < a < 1, then 1/a > 1. Hence, by the first part,
1/a1/n = (1/a)1/n → 1,
so that by Theorem 1.44 (ii), an → 1. ♦
Example 1.64. Let (an) be a bounded sequence of non-negative real numbers. Then (1+an)1/n → 1
as n→∞:
Let M > 0 be such that 0 ≤ an ≤M for all n ∈ N. Then,
1 ≤ (1 + an)1/n ≤ (1 +M)1/n ∀n ∈ N.
Using the result in Example 1.63, (1 +M)1/n → 1. Hence, by the Sandwich theorem (Theorem 1.25
(iv)), (1 + an)1/n → 1. ♦
22
Example 1.65. Let (an) be a sequence of nonnegative terms such that 1 ≤ an ≤ n for all n ∈ N.
Then (1 + an)1/n → 1: Note that
1 ≤ (1 + an)1/n ≤ (1 + n)1/n ≤ (2n)1/n = 21/nn1/n ∀n ∈ N.
By the results in Examples 1.62 and 1.63, 21/n → 1 and n1/n → 1. Hence, by the Sandwich theorem
(Theorem 1.25 (iv)), (1 + an)1/n → 1. ♦
Example 1.66. Let an = (n!)1/n2
, n ∈ N. Then an → 1 as n→∞. We give two proofs for this.
(i) Note that, for every n ∈ N,
1 ≤ (n!)1/n2
≤ (nn)1/n2
= n1/n.
Since n1/n → 1, by the Sandwich theorem (Theorem 1.25 (iv)), we have (n!)1/n2 → 1.
(ii) By GM-AM inequality, for n ∈ N,
(n!)1/n = (1.2. . . . .n)1/n ≤ 1 + 2 + . . .+ n
n=n+ 1
2≤ n.
Thus, 1 ≤ (n!)1/n2 ≤ n1/n. Since n1/n → 1, by the Sandwich theorem (Theorem 1.25 (iv)), we have
(n!)1/n2 → 1. ♦
What about the convergence of the sequence (an) with an = (n!)1/n?
Example 1.67. Let an = (n!)1/n, n ∈ N. We show that an 6→ 1. In fact, we show that (an) is
unbounded.
Note that, for any k, n ∈ N with n ≥ k,
(n!)1/n ≥ (k!)1/n(kn−k)1/n = (k!)1/nk1−k/n = k( k!
kk
)1/n.
Since for any fixed k, k(k!kk
)1/n→ k as n → ∞, we can conclude that (n!)1/n 6→ 1. Now, for k ∈ N,
let nk ∈ N be such that nk ≥ k and ( k!
kk
)1/n≥ 1
2∀n ≥ nk.
Thus,
(n!)1/n ≥ k
2∀ n ≥ nk.
Therefore, the sequence((n!)1/n
)is unbounded. In fact (n!)1/n →∞ as n→∞. ♦
Remark 1.68. Suppose for each k ∈ N, a(k)n → 0, b
(k)n → 1 as n → ∞, and also a
(n)n → 0, b
(n)n → 1
as n→∞. In view of Theorems 1.25 and 1.44, one may think that
a(1)n + a(2)n + · · ·+ a(n)n → 0 as n→∞
23
and
b(1)n b(2)n · · · b(n)n → 1 as n→∞.
Unfortunately, that is not the case. To see this consider
a(k)n =k
n2, b(k)n = k1/n k, n ∈ N.
Then, for each k ∈ N, we have
a(k)n → 0, b(k)n → 1 as n→∞.
Also
a(n)n → 0, b(n)n → 1 as n→∞.
But,
a(1)n + a(2)n + · · ·+ a(n)n =1
n2+
2
n2+ · · ·+ n
n2=n+ 1
2n→ 1
2as n→∞
and from Example 1.67,
b(1)n b(2)n · · · b(n)n = 11/n21/n · · ·n1/n = (n!)1/n 6→ 1 as n→∞. ♦
1.6 Cauchy criterion
Let us observe the following necessary condition for convergent sequences.
THEOREM 1.69. If (an) converges, then for every ε > 0, there exists N ∈ N such that
|an − am| < ε ∀n,m ≥ N.
Proof. Suppose an → a as n → ∞, and let ε > 0 be given. Then we know that there exists N ∈ Nsuch that |an − a| < ε/2 for all n ≥ N . Hence, we have
|an − am| ≤ |an − a|+ |a− am| < ε ∀n,m ≥ N.
This completes the proof.
Is the condition in Theorem 1.69 sufficient as well? The answer is in the affirmative, which is the
so called Cauchy criterion5.
Definition 1.70. A a sequence (an) is said to be a Cauchy sequence if for every ε > 0, there exists
N ∈ N such that
|an − am| < ε ∀n,m ≥ N.
♦5Augustin-Louis Cauchy (21 August 1789 – 23 May 1857) was a French mathematician who made many contributions
to calculus, specifically in terms of its rigorous foundation.
24
THEOREM 1.71. (Cauchy criterion) Every Cauchy sequence of real numbers converges.
Exercise 1.72. Let sn = 1 + 12 + . . .+ 1
n for n ∈ N. Show that |s2n − sn| ≥ 12 . Conclude that (sn) is
a divergent sequence. ♦
Exercise 1.73. Let (an) be a sequence of real numbers. Suppose there exists a positive real number
r < 1 such that
|an+2 − an+1| ≤ r|an+1 − an| ∀ n ∈ N.
Show that (an) is a Cauchy sequence. If a = limn→∞
an, then show that
|a− an| ≤rn−1
1− r|a2 − a1| ∀n ∈ N.
♦
2 Series of Real Numbers
Definition 2.1. A series of real numbers is an expression of the form
a1 + a2 + a3 + . . . ,
or more compactly∞∑n=1
an,
where (an) is a sequence of real numbers. The number an is called the nth term of the series and the
sum of the first n terms of (an), that is,
sn := a1 + · · ·+ an,
is called the nth partial sum of the series. ♦
Remark 2.2. Some authors define a series as the sequence (sn), where sn is nth partial sum of
another sequence (an). ♦
Example 2.3. Following are some examples of series:
(i) 310 + 3
102 + · · · ,
(ii) 1 + 12 + 1
3 + · · · ,
(iii) 1 + 122 + 1
32 + · · ·
Note that in (i), (ii), (iii) above the partial sums are
n∑k=1
3
10k,
n∑k=1
1
k,
n∑k=1
1
k2,
respectively. ♦
25
2.1 Convergence and divergence of series
Definition 2.4. A series∑∞n=1 an is said to be a convergent series if the corresponding sequence
(sn) of partial sums converge. If sn → s, then we say that the series∑∞n=1 an converges to s, and
we write this fact as
s =
∞∑n=1
an.
A series which does not converge is called a divergent series. ♦
Observe the following:
Suppose an ≥ 0 for all n ∈ N. Then the sequence (sn) of partial sums of the series∑∞n=1 an
is monotonically increasing. Hence, in this case, either (sn) converges or sn →∞.
Exercise 2.5. Show that the geometric series
1 + q + q2 + · · ·
for q ∈ R converges if and only if |q| < 1. ♦
THEOREM 2.6. (A necessary condition) If∑∞n=1 an converges, then an → 0 as n → ∞. The
converse is not true.
Proof. Suppose the series∑∞n=1 an converges to s, that is, sn → s, where sn is the nth partial sum of
the series. Since an = sn − sn−1 for n = 2, 3, . . ., we have
an = sn − sn−1 → 0 as n→∞.
Thus, an → 0.
To see that the converse does not hold, consider the series∑∞n=1 1/n. In this case, an = 1/n→ 0,
but the series diverges.
Exercise 2.7. Prove the following.
(i) The series∑∞n=1
nn+1 diverges.
(ii) If an+1 > an > 0 for all n ∈ N, then∑∞n=1 an diverges. ♦
The following theorem shows that if we remove the first few terms from a convergent (respectively,
divergent) series, then the resulting series remain convergent (respectively, divergent).
THEOREM 2.8. For each k ∈ N, the series∑∞n=1 an converges if and only if the series
∑∞n=1 ak+n
converges, and in that case∞∑n=1
an = (a1 + · · ·+ ak) +
∞∑n=1
ak+n.
26
The above theorem shows that if∑∞n=1 bn is obtained from
∑∞n=1 an by omitting or adding a finite
number of terms, then∞∑n=1
an converges ⇐⇒∞∑n=1
bn converges.
The following theorem is an immediate corollary to Theorem 2.8.
THEOREM 2.9. Suppose (an) and (bn) are sequences such that for some k ∈ N, an = bn for all
n ≥ k. Then∑∞n=1 an converges if and only if
∑∞n=1 bn converges.
THEOREM 2.10. Suppose∑∞n=1 an converges to s and
∑∞n=1 bn converges to s′. Then for any
α, β ∈ R,∞∑n=1
(αan + β bn) converges to α s+ β s′.
In particular,∑∞n=1(an + bn) converges to s+ s′ and
∑∞n=1 c an converges to cs for any c ∈ R.
Proof. Let s(1)n , s
(2)n and sn be the nth partial sums of the series
∑∞n=1 an,
∑∞n=1 bn and
∑∞n=1(αan+
β bn) respectively. Then we obtain
sn =
n∑k=1
(αak + β bk) = α
n∑k=1
ak + β
n∑k=1
β bk = αs(1)n + βs(2)n ∀n ∈ N.
Since s(1)n → s and s
(2)n → s′, we obtain sn = αs
(1)n + βs
(2)n → αs+ βs′.
Exercise 2.11. As proof of Theorem 2.10, suppose we write
∞∑n=1
(αan + β bn) = α
∞∑n=1
αan + β
∞∑n=1
bn = αs+ βs′.
What is wrong with it? ♦
2.2 Some tests for convergence
THEOREM 2.12. (Comparison test) Let 0 ≤ an ≤ bn for all n ∈ N.
(i) If
∞∑n=1
bn converges, then
∞∑n=1
an converges.
(ii) If
∞∑n=1
an diverges, then
∞∑n=1
bn diverges.
Proof. Suppose sn and s′n are the nth partial sums of the series∑∞n=1 an and
∑∞n=1 bn respectively.
By the assumption, we have 0 ≤ sn ≤ s′n for all n ∈ N, and both (sn) and (s′n) are monotonically
increasing.
27
(i) Suppose∑∞n=1 bn converges, that is, (s′n) converges. Then, (s′n) is bounded. Hence, by the
relation 0 ≤ sn ≤ s′n for all n ∈ N, (sn) is bounded as well as monotonically increasing. Therefore, by
Theorem 1.50, (sn) converges.
(ii) This part is a restatement of (i).
Exercise 2.13. Suppose (an) and (bn) are sequences of positive terms. Prove the following.
(i) Suppose ` := limn→∞
anbn
exists.
(a) If ` > 0, then∑∞n=1 bn converges ⇐⇒
∑∞n=1 an converges.
(b) If ` = 0, then∑∞n=1 bn converges =⇒
∑∞n=1 an converges.
(ii) Suppose limn→∞
anbn
=∞. Then∑∞n=1 an converges =⇒
∑∞n=1 bn converges.
♦
Example 2.14. We have seen that the sequence (sn) with sn =∑nk=1
1k! converges. This also follows
from comparison test, since1
n!≤ 1
2n−1∀n ∈ N
and∑∞n=1
12n−1 converges. ♦
Example 2.15. (i) Since 1√n≥ 1
n for all n ∈ N, and since the series∑∞n=1
1n diverges, by comparison
test, the series∑∞n=1
1√n
also diverges. More generally, let p ≤ 1. Since
1
np≥ 1
n∀n ∈ N,
by comparison test,∞∑n=1
1
npdiverges for p ≤ 1.
(ii) Let p ≥ 2. We know that∑∞k=1
1n2 converges (Example 1.53). Since 1
np ≤ 1n2 for all n ∈ N,
by comparison test, the series∞∑n=1
1
npconverges for p ≥ 2. ♦
What about the convergence of
∞∑n=1
1
npfor any p > 1? It is dealt in the next example.
Example 2.16. In this example, we shall use the familiarity of the integral to the extent that the
reader knows the result: ∫ b
a
xkdx =
[xk+1
k + 1
]ba
=bk+1 − ak+1
k + 1for k 6= 1.
28
Consider the series∑∞n=1
1np for p > 1. To discuss this general case, consider the function
f(x) := 1/xp, x ≥ 1.
Note that, for each k ∈ N, we have
k − 1 ≤ x ≤ k =⇒ 1
kp≤ 1
xp=⇒ 1
kp≤∫ k
k−1
dx
xp.
Hence,n∑k=2
1
kp≤
n∑k=2
∫ k
k−1
dx
xp=
∫ n
1
dx
xp=n1−p − 1
1− p≤ 1
p− 1.
Thus,
sn :=
n∑k=1
1
kp≤ 1
p− 1+ 1 =
p
p− 1.
Hence, (sn) is monotonically increasing and bounded above, and hence (sn) converges. Thus,
∞∑n=1
1
npconverges for p > 1.
A more general result on convergence of series in terms of integrals, known as integral test, will be
considered later. ♦
Another consequence of the comparison test is the following.
THEOREM 2.17. Let (an) be a sequence of positive numbers. If there exists r with 0 < r < 1 and
a positive integer N such that either
an+1
an≤ r ∀n ≥ N or a1/nn ≤ r ∀n ≥ N,
then the series∑∞n=1 an converges.
Proof. Let 0 < r < 1. First, let N ∈ N be such that
an+1
an≤ r ∀n ≥ N.
Then an+1 ≤ ran for all n ≥ N so that
an+1 ≤ rn−N+1aN =( aNrN−1
)rn ∀n ≥ N.
Since∑∞n=1 r
n converges, the comparison test shows that, the series∑∞n=1 an converges.
Next, let N ∈ N be such that
a1/nn ≤ r ∀n ≥ N.
Then an ≤ rn for all n ≥ N . Again, since∑∞n=1 r
n converges, by comparison test,∑∞n=1 an converges.
29
Sometimes it may be easier to find the limit of either(an+1
an
)or (a
1/nn ) rather than finding an
upper bound. In such cases, the following two tests due to d’Alembert6 and Cauchy, respectively, are
useful to decide the convergence of a series.
THEOREM 2.18. (D’Alembert’s ratio test) Suppose (an) is a sequence of positive terms such that
limn→∞
an+1
an= ` exists.
(i) If ` < 1, then the series∑∞n=1 an converges.
(ii) If ` > 1, then an 6→ 0; in particular,∑∞n=1 an diverges.
Proof. (i) Suppose ` < 1. Let r be such that ` < r < 1. Then there exists N ∈ N such that
an+1
an≤ r ∀ n ≥ N.
Hence, by Theorem 2.17, the series∑∞n=1 an converges.
(ii) Let ` > 1. Then there exists N ∈ N such that
an+1
an> 1 ∀n ≥ N.
Hence, an+1 > an for all n ≥ N . Therefore, an 6→ 0.
THEOREM 2.19. (Cauchy’s root test) Suppose (an) is a sequence of positive terms such that
limn→∞
an1/n = ` exists.
(i) If ` < 1, then the series∑∞n=1 an converges.
(ii) If ` > 1, then an 6→ 0; in particular,∑∞n=1 an diverges.
Proof. (i) Suppose ` < 1. Let r be such that ` < r < 1. Then there exists N ∈ N such that
an1/n ≤ r ∀ n ≥ N.
Hence, by Theorem 2.17, the series∑∞n=1 an converges.
(ii) Let ` > 1. Then there exists N ∈ N such that
an1/n > 1 ∀ n ≥ N.
Hence, an ≥ 1 for all n ≥ N . Therefore, an 6→ 0.
6Jean-Baptiste le Rond d’Alembert (16 November 1717 – 29 October 1783) was a French mathematician, physicist
and philosopher. A particular method of solution of wave equation is named after him - curtsey Wikipedia.
30
Remark 2.20. We remark that both d’Alembert’s ratio test and Cauchy’s root test are silent for the
case ` = 1. In this case, we cannot assert either way. For example, we know that the series∑∞n=1
1n
diverges whereas∑∞n=1
1n2 , and in both the cases, we have
an+1
an→ 1, a1/nn → 1.
However, for such cases, we may be able to infer the convergence or divergence by some other means.
♦
Example 2.21. Let us test the convergence of the series
∞∑n=1
n2
2nand
∞∑n=1
n!
2n.
(i)∞∑n=1
n2
2n: In this case an = n2
2n so that
an+1
an=
(n+ 1)2/2n+1
n2/2n=
1
2
(n+ 1
n
)2→ 1
2.
Hence, by d’Alembert’s ratio test, the series converges.
(ii)
∞∑n=1
n!
2n: In this case an = n!
2n so that
an+1
an=
(n+ 1)!/2n+1
n!/2n=n+ 1
2→∞.
Hence, by d’Alembert’s ratio test, the series diverges. ♦
Example 2.22. For x ≥ 0, the series
∞∑n=1
xn
n!converges:
Clearly the series converges if x = 0. For x > 00, let an = xn
n! . Then we have
an+1
an=
x
n+ 1→ 0.
Hence, by d’Alembert’s ratio test, the series converges. We shall also show that the series
∞∑n=1
xn
n!
converges for any x ∈ R. ♦
Example 2.23. The series
∞∑n=1
( n
2n+ 1
)nconverges: In this case, we have
an1/n =
n
2n+ 1→ 1
2< 1.
31
Hence, by Cauchy’s root test, the series converges.
The convergence of the above series can also be proved using comparison test, since
an =( n
2n+ 1
)n=( 1
2 + 1/n
)n≤ 1
2n
for all n ∈ N and
∞∑n=1
1
2nconverges. ♦
Example 2.24. Consider the series
∞∑n=1
1
n(n+ 1). In this case, we have
limn→∞
an+1
an= 1 = lim
n→∞an
1/n.
Hence, we are not in a position to apply ratio test and root test. However,
sn :=n∑k=1
1
k(k + 1)=
n∑k=1
(1
k− 1
k + 1
)= 1− 1
n+ 1→ 1.
Thus, the series converges to 1. ♦
Exercise 2.25. Assert the convergence of the series∑∞n=1 an in Example 2.21(i) and Example 2.23
by showing thatan+1
an≤ 8
9∀n ≥ 3 and a1/nn ≤ 1
2∀n ∈ N,
respectively. ♦
Exercise 2.26. Assert the divergence of the series∑∞n=1 an in Example 2.21(ii) by showing either
an+1
an≥ 1 ∀n ∈ N or an →∞. /
Exercise 2.27. Show the convergence of the series∑∞n=1
1n2 by comparing it with
∑∞n=1
1n(n+1) . ♦
Example 2.28. Consider the series∑∞n=1 an with
a2n−1 =1
6n−1, a2n =
1
3× 6n−1
for n ∈ N. Note thata2na2n−1
=1
3,
a2n+1
a2n=
1
2
for all n ∈ N so that limn→∞
an+1
andoes not exist. However, an+1
an≤ 1/2 for all n ∈ N so that by Theorem
2.17(i), the series∑∞n=1 an converges. ♦
32
Example 2.29. Consider the series∑∞n=1 an with
an =
( n
2n+ 1
)n, n odd,
1
3n, n even.
Note that
limn→∞
a1/(2n+1)2n+1 = lim
n→∞
2n+ 1
4n+ 3=
1
2, lim
n→∞a1/2n2n =
1
3
so that limn→∞
a1/nn does not exist. However, a1/nn ≤ 1/2 for all n ∈ N. Hence, by Theorem 2.17(i), the
series∑∞n=1 an converges. ♦
2.3 Alternating series
In the last subsection we have described some tests for asserting the convergence or divergence of
series of non-negative terms. In this subsection we provide a sufficient condition for convergence of
those series with alternatively positive and negative terms.
Definition 2.30. A series of the form∑∞n=1(−1)n+1un, where (un) is a sequence of positive terms,
is called an alternating series. ♦
Thus,
1− 1
2+
1
3− 1
4+
1
5+ · · ·+ (−1)n+1
n+ · · ·
is an alternating series, whereas the series
1 +1
2− 1
3− 1
4+ · · ·+ 1
4n− 3+
1
4n− 2− 1
4n− 1− 1
4n+ · · ·
is not an alternating series. As you can see, the above series, though not an alternating series, is of
the form∞∑n=1
[(−1)n+1un + (−1)n+1vn]
with
un =1
2n− 1, vn =
1
2n.
Note that∑∞n=1(−1)n+1un and
∑∞n=1(−1)n+1vn are alternating series. Thus, once we know the
convergence of both the series∑∞n=1(−1)n+1un and
∑∞n=1(−1)n+1vn, we can assert the convergence
of the original series∑∞n=1[(−1)n+1un + (−1)n+1vn].
In view of the above discussion, it is important to have a sufficient condition for the convergence
of alternating series. Following theorem due to Leibnitz7 provides such a condition.
7Gottfried Wilhelm von Leibnitz (July 1, 1646 – November 14, 1716) was a German mathematician and philosopher.
He developed the infinitesimal calculus independently of Isaac Newton - curtsey Wikipedia.
33
THEOREM 2.31. (Leibnitz’s theorem) Suppose (un) is a sequence of positive terms such that
un ≥ un+1 for all n ∈ N and un → 0 as n → ∞. Then the alternating series∑∞n=1(−1)n+1un
converges, and in that case,
|s− sn| ≤ un+1 ∀ n ∈ N,
where
sn =
n∑j=1
(−1)j+1uj and s =
∞∑n=1
(−1)n+1un,
Proof. We observe that
s2n+1 = s2n + u2n+1 ∀ n ∈ N.
Since un → 0 as n→∞, it is enough to show that (s2n) converges, and in that case, both (s2n) and
(s2n+1) converge to the same limit. Note that, for every n ∈ N,
s2n+2 = s2n + (u2n+1 − u2n+2) ≥ s2n
and
s2n = u1 − (u2 − u3)− . . . (u2n−2 − u2n−1)− u2n ≤ u1.
Hence, (s2n) is monotonically increasing and bounded above. Therefore (s2n) converges.
To obtain the remaining part, note also that, for n ∈ N,
s2n+1 = s2n−1 − (u2n − u2n+1) ≤ s2n−1,
so that {s2n−1} is a monotonically decreasing sequence. Thus,
s2n−1 = s2n + u2n ≤ s+ u2n, s ≤ s2n+1 = s2n + u2n+1,
s ≤ s2n−1, s2n ≤ s.
Hence,
s2n−1 − s ≤ u2n, s− s2n ≤ u2n+1,
Consequently,
|s− sn| ≤ un+1 ∀ n ∈ N.
This completes the proof.
COROLLARY 2.32. Suppose (an) is a sequence of positive terms such that an ≥ an+1 for all n ∈ Nand an → 0 as n→∞. Then the series
a1 + a2 − a3 − a4 + · · ·+ a4n−3 + a4n−2 − a4n−1 − a4n + · · ·
converges.
34
Proof. Write the given series as
∞∑n=1
[(−1)n+1un + (−1)n+1vn],
where (un) and (vn) are decreasing sequences of positive terms such that un → 0 and vn → 0, and
then apply Theorem 2.31.
Example 2.33. By Theorem 2.31,
1− 1
2+
1
3− 1
4+ · · ·+ (−1)n+1
n+ · · · ,
1− 1
3+
1
5− 1
7+ · · ·+ (−1)n+1
2n− 1+ · · · ,
1
2− 1
4+
1
6− 1
8+ · · ·+ (−1)n+1
2n+ · · · ,
are convergent series. Using Theorem 2.31, it can also be shown that the series
1 +1
2− 1
3− 1
4+ · · ·+ (−1)n+1
2n− 1+
(−1)n+1
2n+ · · ·
is also a convergent series. ♦
Remark 2.34. The convergence of the series
1− 1
3+
1
5− 1
7+ · · ·+ (−1)n+1
2n− 1+ · · · ,
which is generally known as Leibnitz-Gregory series, was known to Indian mathematicians Madhava
and Nilakantha8 as early as in 15-th century. In fact, it was known that the above series converges toπ4 . We shall prove this in a later section. ♦
Remark 2.35. The relation |s − sn| ≤ un+1 in Theorem 2.31 shows the rate of convergence of the
partial sums to the sum of the series. In particular, for a given k ∈ N, if n is large enough such that
un < 1/10k, then first k decimal places of sn and s are the same. ♦
The following example shows that the condition un ≥ un+1 for all n ∈ N in Theorem 2.31 is not
redundant, that is, the conclusion in the theorem need not hold if the above condition is dropped.
Example 2.36. Consider the series
1√3− 1
− 1√3 + 1
+1√
4− 1− 1√
4 + 1+ · · ·+ 1√
n− 1− 1√
n+ 1+ · · ·
8The series∑∞
n=1(−1)n+1
2n−1appear in the work of a Kerala mathematician Madhava around 1425 which was presented
later in the year around 1550 by another Kerala mathematician Nilakantha. The discovery of the above series is normally
attributed to Leibnitz and James Gregory after nearly 300 years of its discovery.
35
which is of the form∑∞n=1(−1)n+1un with
u2n−1 =1√
n+ 2− 1, u2n =
1√n+ 2 + 1
.
Note that, in this case, we have un ≥ 0 for all n ∈ N and un → 0. However, the series diverges. To
see this, we observe that
s2n =
n∑k=1
( 1√n+ 2− 1
− 1√n+ 2 + 1
)=
n∑k=1
2
n+ 1→∞.
Thus, (sn) has a subsequence which diverges. Therefore, (sn) diverges. Note that (un) is not a
decreasing sequence. ♦
2.4 Absolute convergence
We have seen in Example 1.55 that the series∑∞n=1
|x|nn! converges for every x ∈ R. What about the
convergence of the series∑∞n=1
xn
n! ?
Definition 2.37. Let (an) be a sequence of real numbers. Then the series∑∞n=1 an is said to be
(1) absolutely convergent, if∑∞n=1 |an| is convergent,
(2) conditionally convergent, if it converges, but not absolutely.
♦
Example 2.38. (i) the series
∞∑n=1
(−1)n
n2,
∞∑n=1
(−1)n+1
n!,
∞∑n=1
sinn
n2
are absolutely convergent, so also the series
∞∑n=1
an
n!
for any a ∈ R.
(ii) The series∞∑n=1
(−1)n
n,
∞∑n=1
(−1)n
2n− 1,
∞∑n=1
(−1)n
2n
are not absolutely convergent. They are convergent, by Leibnitz theorem. Thus, these series are
conditionally convergent. ♦
THEOREM 2.39. Every absolutely convergent series is convergent. The converse does not hold.
36
Proof. Suppose∑∞n=1 an is an absolutely convergent series. Let sn and s′n be the nth partial sums of
the series∑∞n=1 an and
∑∞n=1 |an| respectively. Then, for n > m, we have
|sn − sm| =∣∣∣ n∑j=m+1
an
∣∣∣ ≤ n∑j=m+1
|an| = |s′n − s′m|.
Since, (s′n) converges, it is a Cauchy sequence. Hence, from the above relation it follows that (sn) is
also a Cauchy sequence. Therefore, by the Cauchy criterion, it converges.
Recall that the series∑∞n=1
1n does not converge. However, by Theorem 2.31,
∑∞n=1
(−1)n+1
n is
convergent. Thus, the converse of the first statement in the theorem does not hold.
In view of Theorem 2.39, the series
∞∑n=1
xn
n!converges for each x ∈ R. We shall prove in the next
chapter that the value of the above sum is same as
limn→∞
(1 +
x
n
)n,
and it is generally denoted by exp(x) or ex.
In view of Theorem 2.39 we can assert the following.
Let (an) be a sequence of nonzero numbers
(i) If limn→∞
∣∣∣an+1
an
∣∣∣ = ` exists and ` < 1,
then∑∞n=1 anx
n converges absolutely.
(ii) If limn→∞
|an|1/n = ` exists and ` < 1,
then∑∞n=1 anx
n converges absolutely.
Given a convergent series∑∞n=1 an of real numbers, consider another series
∑∞n=1 bn obtained by
rearrangement of the terms of∑∞n=1 an, that is, bn = aϕ(n), n ∈ N, where ϕ is a bijection on N. A
natural question one would like to ask is:
Does∑∞n=1 bn converge to the same limit as that of
∑∞n=1 an?
Let us look at the following example.
Example 2.40. Consider the convergent series
∞∑n=1
(−1)n+1
n. At the moment we do not know to what
number it converges. However, let us see if a rearrangement of it converges to a different number. For
this, let us consider the following rearrangement:
1− 1
2− 1
4+
1
3− 1
6− 1
8+ · · ·+ 1
2n− 1− 1
4n− 2− 1
4n+ · · · .
37
Thus, if an = (−1)n+1
n for all n ∈ N, the rearranged series is written as∑∞n=1 bn, with
b3n−2 =1
2n− 1, b3n−1 = − 1
4n− 2, b3n = − 1
4n
for n ∈ N. Let sn and s′n be the nth partial sums of the series∑∞n=1 an and
∑∞n=1 bn respectively.
Then we have
s′3n = 1− 1
2− 1
4+
1
3− 1
6− 1
8+ · · ·+ 1
2n− 1− 1
4n− 2− 1
4n
=(
1− 1
2− 1
4
)+(1
3− 1
6− 1
8
)+ · · ·+
( 1
2n− 1− 1
4n− 2− 1
4n
)=
(1
2− 1
4
)+(1
6− 1
8
)+ · · ·+
( 1
4n− 2− 1
4n
)=
1
2
[(1− 1
2
)+(1
3− 1
4
)+ · · ·+
( 1
2n− 1− 1
2n
)]=
1
2
[1− 1
2+
1
3− 1
4+ · · ·+ 1
2n− 1− 1
2n
]=
1
2s2n.
Also, we have
s′3n+1 = s′3n +1
2n+ 1, s′3n+2 = s′3n +
1
2n+ 1− 1
4n+ 2.
We know that (sn) converges. Let limn→∞
sn = s. Since, an → 0 as n→∞, we obtain
limn→∞
s′3n =s
2, lim
n→∞s′3n+1 =
s
2, lim
n→∞s′3n+2 =
s
2.
Hence, we can infer that s′n → s/2 as n → ∞. Thus, a rearrangement of a convergent alternating
series need not converge to the same limit. ♦
Not only that. Look at the following astonishing result!
THEOREM 2.41. Suppose∑∞n=1 an is a conditionally convergent series. Then, for each s ∈ R,
there exists a sequence obtained by the rearrangement of the terms of∑∞n=1 an such that the rearranged
series converges to s.
Such situation would not occur if the given series is absolutely convergent. More precisely,
THEOREM 2.42. Suppose∑∞n=1 an is an absolutely convergent series, say to s. Then every series
obtained by rearranging the terms of∑∞n=1 an converge to the same number s.
The proofs of the above two theorems (Theorems 2.41 and 2.42) are quite involved, and hence we
omit the proofs. Interested readers may refer Delninger [?] for the proofs.
38