Prove that if ann 1infinity is a Cauchy sequence then there

Prove that if {a_n}_n = 1^infinity is a Cauchy sequence then there exists an N N such that |a_n| lessthanorequalto max{|a_1|, ..., |a_N|} + 1 Forall n N. Conclude from this (without using Theorem 3.22) that a Cauchy sequence is bounded.

Solution

Since a_n is a Cauchy sequence, there exist N such that |a_n- a_m| <1, for all n,m >= N.

In particular |a_n- a_N| <1.

Hence |a_n| <= |a_n - a_N| + |a_N| < 1+ |a_N| < 1+ max { |a_1|,|a_2|,..., |a_N|} for n >= N

Also for j=1, ..., N , |a_j| < 1+ max { |a_1|,|a_2|,..., |a_N|}

So that  |a_n| <= |a_n - a_N| + |a_N| < 1+ |a_N| < 1+ max { |a_1|,|a_2|,..., |a_N|} for all n.

1+ max { |a_1|,|a_2|,..., |a_N|} is a finite number

for all n |a_n| <= |a_n - a_N| + |a_N| < 1+ |a_N| < 1+ max { |a_1|,|a_2|,..., |a_N|},

So that |a_n| is bounded.