daver00
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Homework Statement
My problem is this, you are given that for a sequence the following is true:
\left|x_{n} - x_{n+1}\right| < \frac{1}{4^{n}}
So its an obvious case to prove the sequence is Cauchy.
Homework Equations
So I'll state the stuff just for the sake of it:
\forall\epsilon>0, \exists N\in\mathbb{N}, s.t. \forall m,n>N \Rightarrow |x_n - x_m|<\epsilon
The Attempt at a Solution
So I start by taking the telescoping sum:
|x_n - x_m| = |x_n - x_{n+1} + x_{n+1} + \cdots - x_{m-1} + x_{m-1} - x_{m}|
|x_n - x_m| \leq |x_n - x_{n+1}| + |x_{n+1} -x_{n+2}|+\cdots + |x_{m-1} - x_{m}|
|x_n - x_m| \leq \sum_{k=n}^{m-1} |x_k - x_{k+1}|
|x_n - x_m| < \sum_{k=n}^{m-1} \frac{1}{4^k} = \frac{4}{3}\left(\frac{1}{4^m} - \frac{1}{4^n}\right)
(I'm hoping this is all correct)
And here I'm stuck, I'm unsure about how to properly select an N and \epsilon ??
Do I simply say something like this:
|x_N - x_{N+K}| < \frac{4}{3}\left(\frac{1}{4^{N+K}} - \frac{1}{4^N}\right)
for some integer K? And then sort of solve for an epsilon? Or should I find something slightly simpler that is also smaller than this and again find some N in this fashion? I'm just confused because I've got three unknowns here, m, n and epsilon, but I need to reduce this to one N right?
I'm kind of new to all this and I'm not really all that cosy on any of it yet, at least I got the telescoping sum part I guess.