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Question

A sequence $\{a_n\}$ of real numbers is said to be a Cauchy sequence of for each $\epsilon$ > 0 there exists a number $N > 0$ such that m, $n > N$ implies that $|a_n − a_m| <\epsilon$.

Prove that every convergent sequence is a Cauchy sequence


Attempt

This is my first time hearing what a cauchy sequence is. I have no idea how to even start this. I googled cauchy sequence and I think its when $a_n$ converges to $a_{n+1}$?

Attempt:

WTS: $\exists a_m \in \mathbb R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in \mathbb N$, if $n > N$, then $|a_n - a_m| < \epsilon$

Let $\epsilon > 0$ be arbitrary

Choose N such that for $n > N$ we have $|a_n - a_m| < \epsilon$

Suppose $n > N$, then

??

Could someone point me to the right direction? Thx.

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  • $\begingroup$ Hi OP! So, in order to prove this, you will need the following fact: "every bounded sequence has a convergent subsequence." Then you will need to show that every Cauchy sequence is bounded - this part is easier, and if you use the definition you should be able to prove it without much trouble. $\endgroup$
    – Chris
    Commented Mar 22, 2017 at 5:27
  • $\begingroup$ The only way I know how to show bounded is by induction and I would need a lowerbound or upperbound to do that too. $\endgroup$
    – user349557
    Commented Mar 22, 2017 at 5:29
  • $\begingroup$ A note: a Cauchy sequence isn't one in which "$a_n$ converges to $a_{n+1}$" (this is also not great phrasing) - it's one in which $a_n$ and $a_{n+m}$ will be $\epsilon$-close when $n > N$, i.e. after a certain index $N$, all the subsequent terms are close together - and don't get any further apart than $\epsilon$. $\endgroup$
    – Chris
    Commented Mar 22, 2017 at 5:30
  • $\begingroup$ Use the Cauchy criterion to bound the "tail" of the sequence, to start. You won't need induction after that! $\endgroup$
    – Chris
    Commented Mar 22, 2017 at 5:32
  • $\begingroup$ Are you implying i do use induction by your last sentence? $\endgroup$
    – user349557
    Commented Mar 22, 2017 at 5:34

4 Answers 4

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The sequence $(a_{n})$ is convergent by assumption; let $l := \lim_{n}a_{n}$. Let $\varepsilon > 0$. Then there is some $N$ such that $|a_{n} - l| < \varepsilon/2$ for all $n \geq N$. Note that $$ |a_{n} - a_{m}| \leq |a_{n}-l| + |l-a_{m}| < \varepsilon/2 + \varepsilon/2 = \varepsilon $$ for all $n,m \geq N$.

The major tool here is the triangle inequality.

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A way to do this is to notice that since the sequence $(x_n)$ is convergent, then a tail of that sequence, let's call it $(x_m)$ is also convergent. That is, for all $\epsilon_1\gt0$, there exists $N\in\mathbb{N}$ such that for all $n\ge N$, $|x_n-x|\lt\epsilon_1$. Similarly, for all $\epsilon_2\gt0$, there exists $N\in\mathbb{N}$ such that for all $m\ge N$, $|x_m-x|\lt\epsilon_2$ for all $m\ge N\in\mathbb{N}$.

Do you know the triangle inequality? If so, you can relate $|x_n-x_m|$, $|x_n-x|\lt\epsilon_1$, and $|x_m-x|\lt\epsilon_2$ such that $\epsilon_1+\epsilon_2=\epsilon$ (a lot of examples you may see in the future use $\epsilon/2$ for these), showing that all convergent sequences are Cauchy.

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The idea behind the standard proof that every convergent sequence is Cauchy is the triangle inequality.

If a sequence $a_n$ converges to a limit $L$ then for every $\epsilon>0$ there is some cutoff point $N$ such that every term past $N$ is within $\epsilon$ of $L$.

This means that any two points in the sequence $a_n,a_m$ with $n,m>N$ are within $\epsilon$ of $L$ so by the triangle inequality they must be within $2\epsilon$ of each other.

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If it is not Cauchy, it is obviously not convergent: for |a_n -a_m| is greater than some epsilon for arbitrarily large n,m. This assures a_n and a_m do not both get within epsilon/2 of some a_infinity.

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