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My understanding from the definition in my book (Rudin) is this.

A seq. $\{p_n\}$ in a metric space $X$ (I only really know $\mathbb R^k$) is said to be a Cauchy sequence if for any given $\epsilon > 0$, $\exists N\in \mathbb N$ such that $\forall n,m\ge N$, $d(p_n,p_m)<\epsilon$.

(1) I see it as, given any tiny value $\epsilon$, we can find a natural number $N$ large enough so that the distance between $p_n$ and $p_m$ is less than $\epsilon$. Am I right ?

The reason I'm asking this is because I was trying to understand the proof of how $$\sum a_nb_n$$ can converge, and the book said this

$$\left\lvert \sum_{n=p}^{q}a_nb_n\right\rvert \leq \epsilon$$

satisfies the Cauchy criterion and therefore it converges.

I read other questions and answers about the Cauchy sequence, but it didn't really help me…

Can someone explain me what's going on?

Edit:

Suppose

a) the partial sums of $A_n = \Sigma a_n$ form a bounded sequence

b) $b_0 \geq b_1 \geq \dotsb$

c) $\lim_{b \to \infty} b_n = 0$

Using the partial summation formula, algebraically the equation in the bottom is proved

$$\left\lvert \sum_{n=p}^{q}a_nb_n \right\rvert \leq \epsilon$$

Algebraically I had no problem, but I don't know why this proves convergence. I thought to show that a sequence is Cauchy, we need to find the distance between two terms in a sequence. That's where I'm confused.

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Since you asked specifically how to understand Cauchy sequences "intuitively" (rather than how to do $\epsilon,\delta$ proofs with them), I would say that the best way to understand them is as Cauchy himself might have understood them. Namely, for all infinite indices $n$ and $m$, the difference $p_n-p_m$ is infinitesmal. Such formalisations exist, for instance, in the context of the hyperreal extension of the field of real numbers.

As far as the particular series you asked about, what is going on is that the book is considering the sequence $p_n$ of partial sums of the series, and applying the Cauchy criterion to this sequence. Then the difference $p_n-p_m$ is the expression $\sum_m^n$ that you wrote down (up to a slight shift in index).

Some thoughts on Cauchy can be found here.

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  • $\begingroup$ Although everyone else helped me very much, I want to note that your answer is the one that I wanted to hear most. Thank you ! $\endgroup$ – hyg17 May 29 '13 at 8:12
  • $\begingroup$ You're welcome. If you are interested in how Cauchy himself might have viewed this, you could consult the article dx.doi.org/10.1007/s10699-011-9235-x $\endgroup$ – Mikhail Katz May 29 '13 at 8:36
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The rough intuition is that if we go far enough along the sequence we get to a point where it doesn't vary very much. And if that is the case it must stay within a narrow range of values.

If we can reduce the variation arbitrarily (choose $\epsilon$ as small as we like) by going far enough ($N$ terms), then we can narrow the range as much as we like, so that there is ultimately a single value - the limit.

The value of the criterion is that it proves there is a limit without needing to know what the limit is - just using the internal properties of the sequence.

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  • $\begingroup$ Thanks a lot for the help ! $\endgroup$ – hyg17 May 29 '13 at 8:11
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It's enough to consider the special case $\mathbb R$ for the purpose of understanding the gist of it. So, we'll assume everything happens in $\mathbb R$.

For a sequence $a_n$ to converge to a finite limit $L$, means that $$\forall \epsilon >0 \exists N s.t. \forall k>N: d(x_k,L)<\epsilon$$ which intuitively means that for any prescribed positive distance $\epsilon$, from some index onwards, all elements in the sequence are within that distance to the limit $L$.

Now, for the sequence to be Cauchy means that $$\forall \epsilon > 0\exists N s.t. \forall n,m>N:d(x_n,x_m)<\epsilon $$ which intuitively means that for any prescribed positive distance $\epsilon$ from some index onwards all elements in the sequence are no more than $\epsilon $ distance from each other.

Before we further examine the crucial differences let us remark that the oh so important property of $\mathbb R$ (and many other spaces) is that they are complete: A sequence converges if, and only if, it satisfies the Cauchy condition.

So, the conditions (in a complete space) actually mean the same thing. So, what are the differences? Well, the condition for convergence to a limit requires you to specify a limit. The Cauchy condition does not, and this is a great thing that occurs often when you want to show something converges but you have no idea what it might converge to. So you show it is Cauchy (in a complete space) and conclude it converges.

Intuitively, it is clear that any convergent sequences is Cauchy. This is so since if all elements from some index onwards are very close to $L$ then they must be very cose to each other as well. The converse though is not clear and not always true (as said above, it is the completeness property of the reals which is not a trivial matter). For instance, in $\mathbb Q$ with the standard metric, any sequence of rational numbers that converges in $\mathbb R$ to an irrational number is Cauchy but fails to converge in $\mathbb Q$.

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  • $\begingroup$ Thank you very much for the detailed explanation between the differences. I understand a lot more now. $\endgroup$ – hyg17 May 29 '13 at 8:10
  • $\begingroup$ you're welcome :) $\endgroup$ – Ittay Weiss May 29 '13 at 8:17
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You are right about you understanding of Cauchy sequence.

Cauchy criterion means that if a real (or complex) sequence is Cauchy, then it is convergent. In this case the sequence is

$ S_{m} = \sum_{n=0}^{m}a_{n}b_{n} $.

Now what is the meaning of $S_{m}$ being Cauchy? And no doubt you would know that the convergence of the series is defined as the convergence of the sequence $S_{m}$.

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  • $\begingroup$ I will edit the question a bit more, so please take a look at it. $\endgroup$ – hyg17 May 29 '13 at 7:47
  • $\begingroup$ Cauchy is not distance between two sequences! It is the difference between the terms of the same sequence. The expression which you said you have no problem in deriving is precisely proves that the sequence $S_{m}$ is Cauchy and as every real (or complex) Cauchy sequence converges, the sequence $S_{m}$ converges. $\endgroup$ – Vishal Gupta May 29 '13 at 8:12
  • $\begingroup$ Sorry, just a miswording. I'm still not used to this, but thanks ! $\endgroup$ – hyg17 May 29 '13 at 8:13
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Let's start from scratch. We say that the infinite series $\sum_{i = 0}^\infty a_i$ converges if the sequence of partial sums $\{S_n\}$ converges, where $S_n := \sum_{i =0}^n a_i$.

Suppose $\{S_n\}$ is a Cauchy sequence. Then by definition, for every $\epsilon > 0$, there exists $N$ such that for all $n, m \ge N$, $|S_m - S_n| \le \epsilon$.

To be more clear, this is equivalent to saying that for all $\epsilon > 0$, there exists $N$ such that $m > n \ge N$ implies

$$\bigg|\sum_{i = 0}^m a_i - \sum_{i=0}^n a_i\bigg| = \bigg|\sum_{i = n + 1}^m a_i \bigg| \le \epsilon$$

which is exactly the Cauchy criterion. Now, it only suffices to show that $\{S_n\}$ was Cauchy to show that $\sum_{i = 0}^\infty a_i$ converges. Why?

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