# Homework Help: Proving a Sequence is Convergent

1. Oct 2, 2008

### bobcat817

1. The problem statement, all variables and given/known data
Let {$$a_{n}$$}$$^{\infty}_{n=1}$$ be a sequence of real numbers that satisfies

|$$a_{n+1}$$ - $$a_{n}$$| $$\leq$$ $$\frac{1}{2}$$|$$a_{n}$$ - $$a_{n-1}$$|

for all n$$\geq$$2

2. Relevant equations

3. The attempt at a solution

So, I know that it suffices to show that the sequence is Cauchy to prove this. And I can show that

|$$a_{m}$$ - $$a_{n}$$| $$\leq$$ $$\sum$$$$^{m-1}_{n}$$$$\frac{1}{2}$$$$^{k-1}$$ * |$$a_{2}$$ - $$a_{1}$$|

And that is about where I get lost. I'm very unclear on how to define the N in relation to the $$\epsilon$$.

2. Oct 2, 2008

I will take your comment that you've shown

$$|a_m - a_n | \le |a_2 - a_1| \sum_{k=n}^m \left(\frac 1 2 \right)^k$$

as correct - make sure it's correct.

Suppose $$\varepsilon > 0$$ is given. Try taking $$N$$ so large that $$n, m$$ larger than $$N$$ give

$$\sum_{k=n}^m \left(\frac 1 2 \right)^k < \frac{\varepsilon}{2|x_2 - x_1|}$$

This should work (why the 2 in my denom - to make sure the sequence terms differ by less than $$\varepsilon$$ - it's a holdover from my student days, and I'm too old to change.)