Cauchy sequences, induction, telescoping property

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SUMMARY

The discussion focuses on proving properties of Cauchy sequences using mathematical induction and the telescoping property. Participants suggest that the first part of the proof can be established through substitution rather than induction. The conversation highlights the importance of understanding the definitions and relationships between terms in the sequence, particularly in proving the general case for k+1. Participants express a need for additional resources and examples to clarify these concepts.

PREREQUISITES
  • Understanding of Cauchy sequences
  • Familiarity with mathematical induction
  • Knowledge of the telescoping property
  • Basic proficiency in sequence notation and manipulation
NEXT STEPS
  • Study the properties of Cauchy sequences in detail
  • Learn about mathematical induction techniques and their applications
  • Explore the telescoping series and its proofs
  • Find textbooks or online resources that provide worked examples of Cauchy sequence proofs
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Students studying real analysis, particularly those focusing on sequences and series, as well as educators seeking to clarify concepts related to Cauchy sequences and induction methods.

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Homework Statement



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

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 textbook 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 |-[tex]\alpha[/tex](b-a)| = [tex]\alpha[/tex]|(b-a)| which I'm a bit dubious about.

Second section of a) let n=1, then:
|s1+1 - s1| = |b - a| = [tex]\alpha[/tex] 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| = [tex]\alpha[/tex]k-1|b - a|
so I think the first part shows the case where k=2 and [tex]\alpha[/tex]k-1=[tex]\alpha[/tex]1=[tex]\alpha[/tex] but I don't know how to make use of this.
??

So moving on to b) sn+1 +[tex]\alpha[/tex]sn where n=1 becomes s2 +[tex]\alpha[/tex]s1 = b + [tex]\alpha[/tex]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. :cry: 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.
 

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You don't need to do induction on the first part of (a). It's just substitution of the formula they provide. If you get stuck at a spot, try going in the reverse direction and finding a spot where it meets the forward direction.

Although that did involve saying |-[tex] \alpha[/tex](b-a)| = [tex] \alpha[/tex]|(b-a)| which I'm a bit dubious about.
That's perfectly fine here since they tell you [tex] \alpha[/tex] > 0.

For the 2nd part of (a), you were correct in assuming it works for k. Now you have to prove it works for k+1. To do this, you'll first have to substitute |s_(k+2) - s_(k+1)| with their respective values using the formula from the first part of (a). Next, after a bit of simplification and rearranging, you should be able to use your induction hypothesis to prove it works for k+1.
 

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