1. ## cauchy

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

2. Hello,
Originally Posted by rmpatel5
If 0<r<1 and for all n with natural numbers, show that is a Cauchy sequence
A sequence $\displaystyle (x_n)$ is a Cauchy sequence if :
$\displaystyle \forall ~ \varepsilon > 0, ~ \exists N \in \mathbb{N}, ~ \text{such that } \forall ~ p,q > N, ~ |x_p-x_q|< \varepsilon$

Note that since $\displaystyle 0<r<1$, $\displaystyle \forall ~ \delta > 0, ~ \exists ~ N \in \mathbb{N}, ~ \text{such that } \forall ~ n > N, ~ r^n < \delta$
because the limit of r^n is 0.

Now, write, assuming p>q, $\displaystyle |x_p-x_q|=|(x_p-x_{p-1})+(x_{p-1}-\dots-x_{q+1})+(x_{q+1}-x_q)|$.

Use the triangle inequality : $\displaystyle |x+y| \le |x|+|y|$

Try to rearrange all this information to answer the question

3. Originally Posted by Moo
Hello,

A sequence $\displaystyle (x_n)$ is a Cauchy sequence if :
$\displaystyle \forall ~ \varepsilon > 0, ~ \exists N \in \mathbb{N}, ~ \text{such that } \forall ~ p,q > N, ~ |x_p-x_q|< \varepsilon$

Note that since $\displaystyle 0<r<1$, $\displaystyle \forall ~ \delta > 0, ~ \exists ~ N \in \mathbb{N}, ~ \text{such that } \forall ~ n > N, ~ r^n < \delta$
because the limit of r^n is 0.

Now, write, assuming p>q, $\displaystyle |x_p-x_q|=|(x_p-x_{p-1})+(x_{p-1}-\dots-x_{q+1})+(x_{q+1}-x_q)|$.

Use the triangle inequality : $\displaystyle |x+y| \le |x|+|y|$

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??