# Cauchy sequences

1. Oct 25, 2004

### tink

Let {an}(n goes from 1 to infinity) be a sequence. For each n define:
sn=Summation(j=1 to n) of aj
tn=Summation(j=1 to n) of the absolute value of aj.

Prove that if
{tn}(n goes from 1 to infinity)
is a Cauchy sequence, then so is
{sn}(n goes from 1 to infinity).

I started this proof with the definition of a Cauchy sequence. Pick an N large enough so that n,m>N makes
|an - am| < epsolon.
So if tn is Cauchy, we have
|tn-tm| < epsolon.
tn-tm = summation|an|-summation|am| = |an|+|an-1|+...+|am+1|
so now
|an| + |an-1| +...+ |am+1| < epsolon
but
|an + an-1 + ... + am+1| < |an|+|an-1|+...+|am+1|
by triangle inequality.
so now
|an + an-1 +...+ am+1| < epsolon
but
|an + an-1 + ... + am+1| = |sn - sm|
so now
|sn-sm| < epsolon, and therefore Cauchy.

Can anybody tell me if this makes sence? Or at least tell me how to write out "summation from n=1 to infinity" on here in symbols??? Thanks so much!

2. Oct 25, 2004

### NateTG

There's a thread on $$\LaTeX$$ somewhere around here...
$$s_n=\sum_{i=0}^{n}a_i$$

The proof looks ok too.

3. Oct 25, 2004

### tink

Thank you SOOO much!