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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!
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!