The Real And Complex Number Systems

∑ log n − x log x x − 1 ≤ log x.
1≤n≤x
In particular, as x ∈ N, we thus have
n log n − n 1 − log n ≤ log n! ≤ n log n − n 1 log n
which implies that
n n−1 e −n1 ≤ n! ≤ n n1 e −n1 .
In each of Exercise 8.9. through 8.14, show that the real-valed sequence a n is
convergent. The given conditions are assumed to hold for all n ≥ 1. In Exercise 8.10
through 8.14, show that a n has the limit L indicated.
8.9 |a n| 2, |a n2 − a n1 | ≤ 1 8 |a 2
n1 − a n2 |.
Proof: Since
|a n2 − a n1 | ≤ 1 8 |a 2
n1 − a n2 |
1 8 |a n1 − a n||a n1 a n|
we know that
So,
≤ 1 2 |a n1 − a n| since |a n| 2
|a n1 − a n| ≤ 1
2
n−1
|a2 − a 1 | ≤ 1
2
n−3
.
k
|a nk − a n| ≤ ∑|a nj − a nj−1 |
j1
k
≤ ∑
j1
1
2
nj−4
≤ 1
2
n−2
→ as n → .
Hence, a n is a Cauchy sequence. So, a n is a convergent sequence.
Remark: (1) If |a n1 − a n| ≤ b n for all n ∈ N, and∑ b n converges, then ∑ a n
converges.
Proof: Since the proof is similar with the Exercise, we omit it.
(2) In (1), the condition ∑ b n converges CANNOT omit. For example,
n 1
(i) Let a n sin ∑ k1
Or
k
(ii) a n is defined as follows:
a 1 1, a 2 1/2, a 3 0, a 4 1/4, a 5 1/2, a 6 3/4, a 7 1, and so on.
8.10 a 1 ≥ 0, a 2 ≥ 0, a n2 a n a n1 1/2 , L a 1 a 22 1/3 .
Proof: If one of a 1 or a 2 is 0, then a n 0 for all n ≥ 2. So, we may assume that
a 1 ≠ 0anda 2 ≠ 0. So, we have a n ≠ 0 for all n. Letb n a n1
a n
, then
b n1 1/ b n for all n
which implies that