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

Section 11.1 Sequences 701
Example 12 The sequence H 3
n 1 5J is decreasing because
The right side is smaller because it has
a larger denominator.
3
n 1 5 . 3
sn 1 1d 1 5 − 3
n 1 6
and so a n . a n11 for all n > 1.
n
Example 13 Show that the sequence a n −
SOLUTION 1 We must show that a n11 , a n , that is,
n 1 1
sn 1 1d 2 1 1 ,
n is decreasing.
n 2 1 1
n
n 2 1 1
This inequality is equivalent to the one we get by cross-multiplication:
n 1 1
sn 1 1d 2 1 1 ,
n
n 2 1 1
&? sn 1 1dsn 2 1 1d , nfsn 1 1d 2 1 1g
&?
n 3 1 n 2 1 n 1 1 , n 3 1 2n 2 1 2n
&?
1 , n 2 1 n
Since n > 1, we know that the inequality n 2 1 n . 1 is true. Therefore a n11 , a n and
so ha n j is decreasing.
SOLUTION 2 Consider the function f sxd −
x
x 2 1 1 :
f 9sxd − x 2 1 1 2 2x 2
sx 2 1 1d 2 − 1 2 x 2
sx 2 1 1d 2 , 0 whenever x 2 . 1
Thus f is decreasing on s1, `d and so f snd . f sn 1 1d. Therefore ha n j is decreasing. n
11 Definition A sequence ha n j is bounded above if there is a number M
such that
a n < M for all n > 1
It is bounded below if there is a number m such that
m < a n for all n > 1
If it is bounded above and below, then ha n j is a bounded sequence.
For instance, the sequence a n − n is bounded below sa n . 0d but not above. The
sequence a n − nysn 1 1d is bounded because 0 , a n , 1 for all n.
We know that not every bounded sequence is convergent [for instance, the sequence
a n − s21d n satisfies 21 < a n < 1 but is divergent from Example 7] and not every
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