Analysis I: Lecture 3

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 (x_n)_{n=1}^{\infty} 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 (x_n)_{n=1}^{\infty} 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 a_1, a_2, a_3, … and b_1, b_2, b_3, … be complex numbers.
    1. If \sum\limits_{j=1}^{\infty} a_j and \sum\limits_{j=1}^{\infty} b_j both converge, then so does \sum\limits_{j=1}^{\infty} (\lambda a_j + \mu b_j) for any complex numbers \lambda and \mu.
    2. ‘Initial terms do not affect convergence.’  That is, if there is some M such that a_j = b_j for j \geq M, then either \sum_{j=1}^{\infty} a_j and \sum_{j=1}^{\infty} b_j 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 (\sum\limits_{j=1}^{n} \frac{1}{j})_{n=1}^{\infty} 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 \sum\limits_{n=1}^{\infty} a_n and \sum\limits_{n=1}^{\infty} b_n, and I tell you that \sum\limits_{n=1}^{\infty} b_n converges, and that 0 \leq a_n \leq b_n for all n \geq 1.  Must the series \sum\limits_{n=1}^{\infty} a_n converge?
  • If I give you a series \sum\limits_{n=1}^{\infty} x_n, and I tell you that \sum\limits_{n=1}^{\infty} |x_n| converges, does that necessarily mean that \sum\limits_{n=1}^{\infty} x_n converges?
  • If I give you a series such that \sum\limits_{n=1}^{\infty} x_n converges, does that necessarily mean that \sum\limits_{n=1}^{\infty} |x_n| converges?
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3 Responses to “Analysis I: Lecture 3”

  1. apgoucher Says:

    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.

  2. Analysis I: Lecture 4 « Theorem of the week Says:

    […] Expositions of interesting mathematical results « Analysis I: Lecture 3 […]

  3. theoremoftheweek Says:

    Thanks, that’s a useful comment. It can be helpful to think about which results would apply to \mathbb{Q} and which really need properties of the real numbers.

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