3.5 no 7 13

Solution :
Let x n be a Cauchy sequence such that x n ∈ Z for every n ∈ N. We want to prove that
x n is ultimately constant, that is
By using Cauchy definition, we know that for every ε > 0, there exists a natural number
H(ε) such that for all natural number n, m ≥ H(ε), the term x n , x m satisfy |x n − x m | < ε.
Now lets use ε = 1. So, there exists a H(1) for n, m ≥ H(1), the term x n , x m satisfy
|x n − x m | < 1.
But since we know that x n ⊂ N, we have
|x n − x m | < 1 ⇒ x n = x m
So, x n becomes ultimately constant for n ≥ H(1).

Solution :
Let x n be a sequence that x 1 = 2 and x n+1 ≔ 2 + 1
for n ≥ 1. We want to prove that
x n
x n is a contractive sequence and find its limit.
We say x n is contractive if there exists a constant C, 0 < C < 1, such that
For all n ∈ N.
In our case, we have :
|x n+2 − x n+1 | ≤ C|x n+1 − x n |
|x n+2 − x n+1 | = |(2 + 1
x n+1
) − (2 + 1 x n
)|
1
= | − 1 |
x n+1 x n
= | x n − x n+1
x n+1 ⋅ x n
|
We need to show that x n ≥ 2 for n ∈ N by using induction.
For n = 1 we have x 1 = 2 ≥ 2
Assume that x n ≥ 2 for arbitrary n ∈ N.
For n + 1 we have :
x n+1 = 2 + 1 x n
≥ 2 + 0
= 2
So, using the mathematical induction, we can conclude that x n ≥ 2 for n ∈ N.

Since we got that x n ≥ 2 for n ∈ N from induction, we have
| x n − x n+1
| ≥ | x n − x n+1
|
x n+1 ⋅ x n 2 ⋅ 2
= 1 4 |x n − x n+1 |
So, with C = 1 , we found that x 4
n is a contractive sequence.
Now, Theorem 3.5.8 states that every contractive is a Cauchy sequence, and therefore is
convergent. Since x n is a contractive sequence, then x n is convergent. Therefore,
lim(x n ) = x exists, and since lim (x n+1 ) = lim(x n ) = x, then
x n+1 = 2 + 1 x n
x n = 2 + 1 x
x 2 = 2x + 1
x 2 − 2x − 1 = 0
(2 ± √4 − 4 ⋅ 1(−1))
x =
2
x a = 1 − √2 < 2
x b = 1 + √2 > 2
Since x n ≥ 2for n ∈ N, we know that x n < 2 can’t be limit, so we can conclude that
lim(x n ) = 1 + √2