# Proving a Sequence is Cauchy

Suppose a sequence $\{a_n\}$ has this property: there exist constants $C$ and $K$, with $0<K<1$ such that $\vert a_n - a_{n+1} \vert < CK^n$, for $n \gg 1$. Prove that $\{a_n\}$ is a Cauchy sequence.

My initial setup was to show that it was bounded by letting $K = \frac{1}{2}$ and then showing that for any $\epsilon$, we have $\vert a_n - a_{n+1} \vert < CK^n$ such that $n<\log_2{\vert\frac{C}{\epsilon}\vert}$. However, my attempts did not really show anything about this being a Cauchy sequence. I know that I need to use the definition of Cauchy sequences, but I do not know how to proceed. Thanks in advance; any help is greatly appreciated since I have been on this problem for some time now.

• Hint: $|a_n-a_m|\le\sum_{j=m}^{n-1}|a_{j+1}-a_j|\le C \sum_{j=m}^{n-1}K^j$ – Pocho la pantera Sep 18 '13 at 1:09
• For $C \sum_{j=m}^{n-1}K^j$, could we use a value of K that lies between the given $0 < K <1$ such that $K$ converges? – Jamil_V Sep 18 '13 at 2:18
• We can! It is a geometric series since we have $0<K<1$. So we should have something like $C \frac{1}{1-K}$ right? – Jamil_V Sep 18 '13 at 2:28
• Something like this. $|a_n-a_m|\le C K^m ...$ – Pocho la pantera Sep 18 '13 at 2:31

Here is a hint to help you get started. First, recall the definition of a Cauchy sequence. We need to show that for each $\epsilon > 0$, there exists an $N$ such that for all $m,n > N$, we have $|a_n-a_m| < \epsilon$.
Let's just consider $|a_n-a_m|$ and assume $m > n$. Then, we can use the triangle inequality to get
$|a_n-a_m| = |a_n -a_{n+1} + a_{n+1}-a_{n+2} - .... + a_{m-1}-a_m| \leq \sum_{i=n}^{m-1} |a_{i+1}-a_i|$
• Use the provided information. You know that each term $|a_{i+1}-a_i| \leq CK^i$. Now, choose an $\epsilon$. then find an $N$ such that for all $n,m > N$, you have $|a_n -a_m| < \epsilon$ using the information we have so far. – user35959 Sep 18 '13 at 4:01