How do I prove the convergence of a Cauchy sequence using a telescoping sum?

[daver00]

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.

 

[Focus]

You don't need to choose an epsilon. You need to pick an N dependent on epsilon. Your n, m aren't really unknowns, you know they are both greater than N. So you just pick an N.

From my experience, in these sorts of questions, you don't need to make explicit what N you choose, what you did is correct, all you need to say is that the infinite sum (of 1/4^k) is finite, which then tells you the tails of the sum tend to zero. So you can pick an N s.t.
$$\displaystyle \sum_{k=n}^{m}\frac{1}{4^k}$$ for n,m>N.

 

[maze]

You're really close! Take a look at what you wrote right here:
$$|x_n - x_m| < \sum_{k=n}^{m-1} \frac{1}{4^k} = \frac{4}{3}\left(\frac{1}{4^m} - \frac{1}{4^n}\right)$$

if m>N and n>N, can you make a bound on the right hand side? eg:

$$\frac{4}{3}\left(\frac{1}{4^m} - \frac{1}{4^n}\right) \le ??$$

 

[daver00]

So is it just something like this?
$$\frac{4}{3}\left(\frac{1}{4^m} - \frac{1}{4^n}\right) \leq \frac{1}{3\cdot4^N}$$
or even just this:
$$\frac{4}{3}\left(\frac{1}{4^m} - \frac{1}{4^n}\right) \leq \frac{1}{4^N}$$

I'm fairly certain we need to make this stuff pretty water tight in this course, so I'm guessing that once I choose my N I'm going to need to work my way back to show the difference is less than epsilon, or is that just not usually the case for this sort of thing?

 

[maze]

yeah exactly. So now if I give you epsilon = .01, what N must you choose to make |xn - xm|<epsilon?

 

[daver00]

I see! so I make it like this:
$$\frac{1}{4^N} < \epsilon$$

$$\frac{1}{4}\ln\left(\frac{1}{\epsilon}\right) < N$$

Or in other words, to answer you question I choose:
$$N>0.25\ln(100)>1.15$$, so $$N \geq 2$$

Thank you so much, sorry for not being any good at the simple part. I'm a mech engineering student but I've recently started a degree in maths as well and while I can cope with really hard technical things, the simple logic is still strange voodoo to me...