In which we study another notion of convergence of a sequence, and see how it relates to our existing definition.
- Definition of a Cauchy sequence.
- Lemma 4: Let be a convergent sequence of real numbers. Then it is a Cauchy sequence. We saw that this followed easily from the definitions.
- Lemma 5: Let be a Cauchy sequence of real numbers. Then the sequence is convergent. We first proved that the sequence is bounded, then applied the Bolzano-Weierstrass theorem to obtain a convergent subsequence, then showed that in fact the whole sequence must be convergent.
- Definition of convergence of a series.
- Lemma 6: Let , , , … and , , , … be complex numbers.
- If and both converge, then so does for any complex numbers and .
- ‘Initial terms do not affect convergence.’ That is, if there is some such that for , then either and both converge or they both diverge.
To prove (i), we simply used the definition (being careful to work with partial sums). We’ll prove (ii) next time; I encourage you to try to prove it yourself before then.
- I handed out the first examples sheet.
Understanding today’s lecture
You could pick some examples of sequences to check directly whether they converge and whether they are Cauchy sequences. Make sure that you find ‘non-examples’ as well as examples! Is the sequence a Cauchy sequence? What happens if you try to extend the ideas from today’s lecture to the complex numbers?
Further reading
Here’s another relevant Tricki article, this time about proving that a sequence converges by finding a convergent subsequence. Inevitably Wikipedia has something to say on the subject of Cauchy sequences.
Preparation for Lecture 4
- I give you two series and , and I tell you that converges, and that for all . Must the series converge?
- If I give you a series , and I tell you that converges, does that necessarily mean that converges?
- If I give you a series such that converges, does that necessarily mean that converges?
January 24, 2013 at 1:40 am
There’s an explanation for the inherent difficulty in proving that Cauchy sequences converge — the reason being that it’s true in R, but not in Q. For instance, the rational sequence {1, 2, 3/2, 5/3, 8/5, 13/8} is Cauchy in Q, but does not converge in Q (its limit is an irrational number, phi). Hence, proving that a limit exists necessarily requires the Least Upper Bound axiom or one of its corollaries (e.g. Bolzano-Weierstrass, as in the lecture). By comparison, the fact that convergence implies Cauchyness is relatively trivial.
Sam Cappleman-Lynes mentioned that this is an alternative way to construct the reals (the traditional method being Dedekind cuts) — two Cauchy sequences of rationals converge to the same real number if and only if we obtain another Cauchy sequence by interleaving the terms. It’s not too difficult to demonstrate that this relationship is an equivalence relation, and we can identify reals with equivalence classes of Cauchy sequences.
January 25, 2013 at 1:30 pm
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January 30, 2013 at 10:14 am
Thanks, that’s a useful comment. It can be helpful to think about which results would apply to and which really need properties of the real numbers.