Algebra Infinite sequences and series. plus Appendix

CHAPTER 4

Infinite Sequences and Series


4.1. Sequences

A sequence is an infinite ordered list of numbers, for example the
sequence of odd positive integers:
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 . . .
Symbolically the terms of a sequence are represented with indexed
letters:
a1 , a2 , a3 , a4 , a5 , a6 , a7 , . . . , an , . . .
Sometimes we start a sequence with a0 (index zero) instead of a1 .
Notation: the sequence a1 , a2 , a3 , . . . is also denoted by {an } or
{an }∞
n=1 .

Some sequences can be defined with a formula, for instance the
sequence 1, 3, 5, 7, . . . of odd positive integers can be defined with the
formula an = 2n − 1.
A recursive definition consists of defining the next term of a se-
quence as a function of previous terms. For instance the Fibonacci
sequence starts with f1 = 1, f2 = 1, and then each subsequent term is
the sum of the two previous ones: fn = fn−1 +fn−2 ; hence the sequence
is:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .

4.1.1. Limits. The limit of a sequence is the value to which its
terms approach indefinitely as n becomes large. We write that the limit
of a sequence an is L in the following way:
lim an = L or an → L as n → ∞ .
n→∞

For instance
1
lim = 0,
n→∞ n
83

, 4.1. SEQUENCES 84

n+1
lim = 1,
n→∞ n
etc.
If a sequence has a (finite) limit the it is said to be convergent,
otherwise it is divergent.
If the sequence becomes arbitrarily large then we write

lim an = ∞ .
n→∞

For instance
lim n2 = ∞ .
n→∞



4.1.2. Theorem. Let f be a function defined in [1, ∞]. If limx→∞ f (x) =
L and an = f (n) for integer n ≥ 1 then limn→∞ an = L (i.e., we can
replace the limit of a sequence with that of a function.)
ln n
Example: Find lim .
n→∞ n

ln x
Answer : According to the theorem that limit equals lim , where
x→∞ x
x represents a real (rather than integer) variable. But now we can use
L’Hôpital’s Rule:
ln x (ln x)0 1/x
lim = lim = lim = 0,
x→∞ x x→∞ (x) 0 x→∞ 1

hence
ln n
lim =0 .
n→∞ n



Example: Find lim rn (r > 0).
n→∞

Answer : This limit is the same as that of the exponential function
x
r , hence


0 if 0 < r < 1
n
lim r = 1 if r = 1
n→∞ 
∞ if r > 1

, 4.1. SEQUENCES 85

4.1.3. Operations with Limits. If an → a and bn → b then:
(an + bn ) → a + b.
(an − bn ) → a − b.
can → ca for any constant c.
an bn → ab.
an a
→ if b 6= 0.
bn b
(an )p → ap if p > 0 and an > 0 for every n.
n2 + n + 1
Example: Find lim .
n→∞ 2n2 + 3

Answer : We divide by n2 on top and bottom and operate with limits
inside the expression:
n2 + n + 1 1 + n1 + n12 1+0+0 1
lim = lim 3 = = .
n→∞ 2n2 + 3 n→∞ 2 + n2 2+0 2

4.1.4. Squeeze Theorem. If an ≤ bn ≤ cn for every n ≥ n0 and
lim an = lim cn = L, then lim bn = L.
n→∞ n→∞ n→∞

Consequence: If lim |an | = 0 then lim an = 0.
n→∞ n→∞

cos n
Example: Find lim .
n→∞ n
1 cos n 1 1
Answer : We have − ≤ ≤ , and → 0 as n → ∞, hence
n n n n
by the squeeze theorem
cos n
lim =0 .
n→∞ n

4.1.5. Other definitions.
4.1.5.1. Increasing, Decreasing, Monotonic. A sequence is increas-
ing if an+1 > an for every n. It is decreasing if an+1 < an for every n.
It is called monotonic if it is either increasing or decreasing.
n+1
Example: Probe that the sequence an = is decreasing.
n

, 4.1. SEQUENCES 86

n+2 n+1 −1
Answer : an+1 − an = − = < 0, hence
n+1 n n(n + 1)
an+1 < an for all positive n.
4.1.5.2. Bounded. A sequence is bounded above if there is a number
M such that an ≤ M for all n. It is bounded below if there is a number
m such that m ≤ an for all n. It is called just bounded if it is bounded
above and below.
n+1
Example: Probe that the sequence an = is bounded.
n
Answer : It is in fact bounded below because all its terms are posi-
tive: an > 0. To prove that it is bounded above note that
n+1 1
an = = 1 + ≤ 2.
n n
since 1/n ≤ 1 for all positive integer n.

4.1.6. Monotonic Sequence Theorem. Every bounded mono-
tonic sequence is convergent.
n+1
For instance, we proved that an = is bounded and monotonic,
n
n+1
so it must be convergent (in fact n
→ 1 as n → ∞).
Next example shows that sometimes in order to find a limit you
may need to make sure that the limits exists first.
Example: Prove that the following sequence has a limit. Find it:
q r q
√ √ √
2, 2 + 2, 2 + 2 + 2, . . .

√
Answer
√ : The sequence can be defined recursively as a 1 = 2,
an+1 = 2 + an for n ≥ 1. First we will prove by induction that
0 < an < 2, so the sequence is bounded.
√
We start (base of induction) by noticing that 0 < a1 = 2 < 2.
Next the induction step. Assume (induction hypothesis) that for a
given value of n it is true that 0 < an < 2. From here we must prove
that the same is true for the next value of n, i.e. that 0 < a√ n+1 < 2.
2
In fact (an+1 ) = 2 + (an ) < 2 + 2 = 4, hence 0 < an+1 < 4 = 2,
q.e.d. So by the induction principle all terms of the sequence verify
that 0 < an < 2.