# Cauchy sequences, induction, telescoping property

1. Jul 22, 2010

### Meggle

1. The problem statement, all variables and given/known data

Scanned and attached

2. Relevant equations

I am guessing a combination of induction and the telescoping property.

3. The attempt at a solution

I'm studying this extramurally, and I've just hit a wall with this last chunk of the sequences section, so if someone can suggest a good text book that would be really good too. My googlefu is not helping me this week.

Ok. So prove the first chunk of a). The only way I can think of to prove that chunk is by mathematical induction again, and I think from the fact that it doesn't explicitly say to use that that there ought to be another way. But I can't see it. So I think I've proved it using mathematical induction. Although that did involve saying |-$$\alpha$$(b-a)| = $$\alpha$$|(b-a)| which I'm a bit dubious about.

Second section of a) let n=1, then:
|s1+1 - s1| = |b - a| = $$\alpha$$ 0|b - a|
so the result holds for n=1.
Assume the result holds for some positive integer k, i.e. assume:
|sk+1 - sk| = $$\alpha$$k-1|b - a|
so I think the first part shows the case where k=2 and $$\alpha$$k-1=$$\alpha$$1=$$\alpha$$ but I don't know how to make use of this.
??

So moving on to b) sn+1 +$$\alpha$$sn where n=1 becomes s2 +$$\alpha$$s1 = b + $$\alpha$$a ok for n=1. But I don't know how to prove the general, as the method my readings have relies on incorporating the definition of <sn> into the equation to be proved, and I don't have a definition for sk+1, or any examples in my readings of a proof involving sk+2 instead of sk+1.

Hence why I feel like I've just not got the jist of this section of the course material, but I've been back and forth through my readings and it's not becoming any clearer. Can anyone suggest the fundamental thinking I'm obviously not getting? Even some worked examples somewhere would be brilliant.

And I don't know why all my alphas are floating up so high, sorry. :yuck:

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2. Jul 22, 2010

That's perfectly fine here since they tell you $$\alpha$$ > 0.