# Cauchy Sequence

• Oct 19th 2012, 04:10 PM
lovesmath
Cauchy Sequence
I need to prove that every Cauchy sequence is bounded. Here is my work thus far:

{a(sub n)}n=1 to infinity is a Cauchy sequence. A sequence is said to be Cauchy if for all E>0 there exists N in the natural numbers such that if m,n>=N, then |a(sub n)-a (sub m)|<E.
Proof: Let E=1. Then there exists N in the natural numbers such that m,n>=N implies that |a(sub n)-a(sub m)|<1. By the reverse triangle inequality, ||a(sub n)|-|a(sub m)||<=|a(sub n)-a(sub m)|<1 for m,n>=N, so |a(sub n)|<=|a(sub N+1)|+1 for n>=N. (Fix m=N+1). Let M=max{|a(sub 1)|, |a(sub 2)|, ..., |a(sub N+1)|+1} then |a(sub n)|<=M.

Can you tell me if I am missing any information?
• Oct 19th 2012, 04:30 PM
Plato
Re: Cauchy Sequence
Quote:

Originally Posted by lovesmath
I need to prove that every Cauchy sequence is bounded.

Because this may well be a most important theorem in sequences, I think its proof needs to done correctly.

$1>0$ so $\exists N$ such that if $n\ge N$ then $|a_n-a_N|<1$
Of course that means $|a_n|\le 1+|a_N|$ if $n\ge N$

Let $B = 1 + \sum\limits_{k = 1}^N {|a_k |}$.

Thus $|a_n|\le B$ for $\forall n~.$