James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

694 Chapter 11 Infinite Sequences and Series
A sequence can be thought of as a list of numbers written in a definite order:
a 1 , a 2 , a 3 , a 4 , . . . , a n , . . .
The number a 1 is called the first term, a 2 is the second term, and in general a n is the nth
term. We will deal exclusively with infinite sequences and so each term a n will have a
successor a n11 .
Notice that for every positive integer n there is a corresponding number a n and so a
sequence can be defined as a function whose domain is the set of positive integers. But
we usually write a n instead of the function notation f snd for the value of the function at
the number n.
NotatioN The sequence {a 1 , a 2 , a 3 , . . .} is also denoted by
`
ha n j or ha n j n−1
(a) H n
`
n 1 1Jn−1
Example 1 Some sequences can be defined by giving a formula for the nth term. In
the following examples we give three descriptions of the sequence: one by using the
preceding notation, another by using the defining formula, and a third by writing out
the terms of the sequence. Notice that n doesn’t have to start at 1.
a n −
H n
1 n 1 1 2 , 2 3 , 3 4 , 4 5 , . . . , n
J
n 1 1 , . . .
(b) H s21dn sn 1 1dJ a
3 n n − s21dn sn 1 1d
H2 2 3 n 3 , 3 9 , 2 4
27 , 5 81 , . . . , s21dn sn 1 1d
3 n , . . .J
`
(c) hsn 2 3 j n−3
a n − sn 2 3 , n > 3 h0, 1, s2 , s3 , . . . , sn 2 3 , . . .j
`
(d) Hcos n 6
Jn−0
a n − cos n 6 , n > 03
H1, s3
2 , 1 2 , 0, . . . , cos n 6 , . . . J n
Example 2 Find a formula for the general term a n of the sequence
H 3 , 2 4
5 25 , 5
125 , 2 6
625 , 7
J
3125 , . . .
assuming that the pattern of the first few terms continues.
SOLUTION We are given that
a 1 − 3 5
a 2 − 2 4
25
a 3 − 5
125
a 4 − 2 6
625
a 5 − 7
3125
Notice that the numerators of these fractions start with 3 and increase by 1 whenever
we go to the next term. The second term has numerator 4, the third term has numerator
5; in general, the nth term will have numerator n 1 2. The denominators are the
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