sequence

ErIcK★TsE

1.


2.


[ 本帖最後由 ErIcK★TsE 於 2009-8-18 09:56 AM 編輯 ]

Naozumi

1.

Since (an) is a bounded sequence, there is a contant M > 0 s.t.  |an| < M for n = 1, 2,....

Since bn -> 0,
For all epsilon > 0, there is a positive integer N s.t. |bn - 0| < epsilon/M for n > N

so when n > N
|an bn - 0| = |an| |bn| < M |bn| < M* epsilon/M = epsilon

2.
Suppose b > c.
so b - c > 0
since an -> b, there is an integer N1  s.t.  b - an < (b - c)/3  
since an -> c, there is an integer N1  s.t.  an - c < (b - c)/3
Summing up the inequalities, we get b - c < 2(b - c)/3, which is a contradiction