The Real And Complex Number Systems

x n1 x n x n 3 6
x
7 n
x n 3 7x n 6
7
0since0 x n 1 for all n N.
It means that the sequence x n (strictly) increasing. Since x n is bounded, by
completeness of R, we know that he sequence x n is convergent, denote its limit by x.
Since
x
x lim n
x n1 lim n3 6
n
7
x3 6 ,
7
we find that x 3, 1, or 2. Since 0 x n 1 for all n N, we finally have x 1.
Claim that if x 1 3 ,then1 x 2 n 2 for all n N. We prove the claim by
Mathematical Induction. Asn 1, there is nothing to prove. Suppose n k holds, i.e.,
1 x k 2, then as n k 1, we have
1 1 6 x k1 x k 3 6
23 6 2.
7
7 7
Hence, we have prove the claim by Mathematical Induction. Since
x 3 7x 6 x 3x 1x 2, then
x n1 x n x n 3 6
x
7 n
x n 3 7x n 6
7
0since1 x n 2 for all n N.
It means that the sequence x n (strictly) decreasing. Since x n is bounded, by
completeness of R, we know that he sequence x n is convergent, denote its limit by x.
Since
x
x lim n
x n1 lim n3 6
n
7
x3 6 ,
7
we find that x 3, 1, or 2. Since 1 x n 2 for all n N, we finally have x 1.
Claim that if x 1 5 , then x 2 n 5 for all n N. We prove the claim by
2
Mathematical Induction. Asn 1, there is nothing to prove. Suppose n k holds, i.e.,
x k 5 , then as n k 1,
2
x k1 x k 3 6
5 2 3 6
7 7
173
56 3 5 2 .
Hence, we have proved the claim by Mathematical Induction. Ifx n was convergent,
say its limit x. Then the possibilities for x 3, 1, or 2. However, x n 5 for all n N.
2
So, x n diverges if x 1 5 . 2
Remark: Note that in the case x 1 5/2, we can show that x n is increasing by the
same method. So, it implies that x n is unbounded.
4.6 If |a n| 2and|a n2 a n1 | 1 8 |a 2
n1 a n2 | for all n 1, prove that a n
converges.
Proof: Let a n1 a n b n , then we have |b n1 | 1 8 |b n||a n1 a n| 1 2 |b n|, since
|a n| 2 for all n 1. So, we have |b n1 | 1 2 n |b 1 |. Consider (Let m n)