Thread: cauchy

If 0<r<1 and for all n with natural numbers, show that is a Cauchy sequence

Hello,

Originally Posted by rmpatel5

If 0<r<1 and for all n with natural numbers, show that is a Cauchy sequence

A sequence is a Cauchy sequence if :


Note that since ,
because the limit of r^n is 0.

Now, write, assuming p>q, .

Use the triangle inequality :

Try to rearrange all this information to answer the question

Originally Posted by Moo

Hello,

A sequence is a Cauchy sequence if :


Note that since ,
because the limit of r^n is 0.

Now, write, assuming p>q, .

Use the triangle inequality :

Try to rearrange all this information to answer the question

consider | x n+p - xn| = | xn+p - x n+p-1 + xn+p-1 -xn+p-2+xn+p-2-xn+p-3+xn+p-3- -------xn+1+xn+1-xn |
= | xn+p - x n+p-1| + | xn+p-1 -xn+p-2|+ |xn+p-2-xn+p-3||+ | xn+p-3- ------+ |xn+2-xn+1| + | xn+1-xn |
< r n+p-1 + r n+p-2 + rn+p-3 + --- + r n <
we know that as n tends to ∞ , r n tends to 0.
so, | x n+p - xn| < where = ε is as small as required depending on p.
so, | x n+p - xn| < ε for all n,p > m where m depends on ε.
so, { xn} is a cauchy sequence.

right??