# Math Help - Cauchy sequence

1. ## Cauchy sequence

Prove that a sequence converges iff it is a Cauchy sequence.

So the sequence $a_n \to L$. For $\varepsilon > 0$ there is an $N \in \bold{N}$ such that $j \geq N \implies |a_{n}-L| < \varepsilon/2$. And so $m,n \geq N \implies |a_{m}-a_{n}| = |a_{m}-L +L-a_{n}| \leq |a_{m}-L|+|L-a_{n}| < \varepsilon$ (by triangle inequality).

Now suppose $a_n$ is a Cauchy sequence. Choose $\varepsilon = 1/2$. Then there is an $N \in \bold{N}$ such that $n,m \geq N \implies |a_{n}-a_{m}| < 1/2$. So $a_N - 1/2 < a_n < a_N+1/2$.

Is this the correct way of proving this?

2. Originally Posted by particlejohn
Prove that a sequence converges iff it is a Cauchy sequence.
So the sequence $a_n \to L$. For $\varepsilon > 0$ there is an $N \in \bold{N}$ such that $j \geq N \implies |a_{n}-L| < \varepsilon/2$. And so $m,n \geq N \implies |a_{m}-a_{n}| = |a_{m}-L +L-a_{n}| \leq |a_{m}-L|+|L-a_{n}| < \varepsilon$ (by triangle inequality).
That part is certainly correct and well done. What are you doing with the converse?

If $a_n$ is a Cauchy sequence, then you need to find a number to which it converges.
Again I don’t know what theorems you have.

Here is one way to proceed.
1. Every sequence contains a monotone subsequence.
2. Every Cauchy sequence is bounded.
3. Every bounded monotone sequence converges.
4. Prove the limit of that subsequence is the limit of the whole sequence.

3. Originally Posted by particlejohn
Now suppose $a_n$ is a Cauchy sequence. Choose $\varepsilon = 1/2$. Then there is an $N \in \bold{N}$ such that $n,m \geq N \implies |a_{n}-a_{m}| < 1/2$. So $a_N - 1/2 < a_n < a_N+1/2$.
You proved the forward direction correctly. (In fact in any metric space a convergent sequence is Cauchy. However, not necessarily the other way around. When the converse hold we say the metric space is complete).

The way I like to prove this fact is by introducing the notion of limsup's and liminf's. Let $\{a_n\}$ be a sequence. If $\{a_n\}$ is not bounded from above then we define $\limsup a_n = \infty$. Otherwise, we contruct the following sequence: $s_1 = \sup\{ s_k|k\geq 1\} = \sup\{ s_1,s_2,...\}$, $s_2 = \sup\{ s_k|k\geq 2\} = \sup\{ s_2,s_3,...\}$, in general $s_n = \sup\{ s_k | k\geq n\}$. Of couse these sequences make sense because we assumed the sequence was bounded from above. Let us define $s_n$'s to be the superior sequence. Note that $s_1\geq s_2\geq s_3 \geq ...$. Thus, $\{ s_n\}$ is a non-increasing sequence and it must have a limit (or $-\infty$). Now we define $\limsup a_n = \lim s_n$ this is the limsup of $a_n$. The beauty of this is that the sequence $a_n$ does not need to have a limit, not even $\pm \infty$ divergence, but yet the limit superior always exists. Now we can define limit inferior in an analogus fashion. If $\{a_n\}$ is not bounded from below we define $\liminf a_n = -\infty$. Otherwise construct $i_n = \inf\{ s_k | k\geq n\}$ just like above. Now confirm that $i_1 \leq i_2 \leq i_3 \leq ...$. This means $\{ i_n\}$ the inferior sequence is a non-decreasing sequence. Thus, it must have a limit (or $+\infty$). We define $\liminf a_n = \lim i_n$. Finally if $\{a_n\}$ is bounded (meaning both from above and below) then $s_n$ has a lower bound (i.e. $i_1$) which automatically means $\limsup a_n$ is a real number. Likewise if $\{a_n\}$ is bounded then $i_n$ has an upper bound (i.e. $s_1$) which automatically means $\liminf a_n$ is a real number.
There are a number of nice properties that are not hard to prove, let $L = \mathbb{R}\cup \{\pm \infty\}$:
• if $\lim a_n = L$ then $\liminf a_n = \limsup a_n = L$
• if $\lim a_n = L$ then $\liminf a_n \leq L \leq \limsup a_n$
• $\liminf a_n \leq \limsup a_n$
• if $\liminf a_n = \limsup a_n = L$ then $\lim a_n = L$

We are now able to elegantly prove that Cauchy sequences are convergent. The first fact we will need is that is $\{a_n\}$ is a Cauchy sequence then $\{a_n\}$ is a bounded sequence. I leave that for you to prove, it happens to be easy. Since $a_n$ is bounded it means $\liminf a_n$ and $\limsup a_n$ are real numbers. To show $\{a_n\}$ is convergent is sufficies to use property # 4 from that list. Let $\epsilon > 0$. Since $\{a_n\}$ is Cauchy it means there is $N\in \mathbb{N}$ such that $|a_n - a_m | < \epsilon$ for all $n,m\geq N$. Thus, $a_m - \epsilon < a_n < a_m + \epsilon$ for all $n,m\geq N$. Thus, $a_m + \epsilon$ is an upper bound for $a_n$, so $s_N = \sup\{ s_k | k\geq N\} \leq a_m + \epsilon$. But this show $s_N - \epsilon < a_m$ i.e. $s_N - \epsilon$ is a lower bound for $a_m$. Thus, $s_N - \epsilon \leq \inf \{ a_k | k\geq N \} = i_N$. This means $\limsup a_n \leq s_N \leq i_N + \epsilon \leq \liminf a_n + \epsilon$. This statement is true for any $\epsilon > 0$ so $\limsup a_n \leq \liminf a_n$. But by property #3 we have $\liminf a_n \leq \limsup a_n$. Thus, $\liminf a_n = \limsup a_n$. This prove that $\{a_n\}$ is convergent.

The reason why limsup's and liminf's are helpful in proving convergence is that they are always convergent if the sequence is bounded. And so dealing with them is always possible.

4. Originally Posted by Plato
That part is certainly correct and well done. What are you doing with the converse?

If $a_n$ is a Cauchy sequence, then you need to find a number to which it converges.
Again I don’t know what theorems you have.

Here is one way to proceed.
1. Every sequence contains a monotone subsequence.
2. Every Cauchy sequence is bounded.
3. Every bounded monotone sequence converges.
4. Prove the limit of that subsequence is the limit of the whole sequence.

Now suppose $a_n$ is a Cauchy sequence. Choose $\varepsilon = 1/2$. Then there is an $N \in \bold{N}$ such that $n,m \geq N \implies |a_{n}-a_{m}| < 1/2$. So $a_N - 1/2 < a_n < a_N+1/2$ for all $n \geq N$.

So $\{a_{n}: n \geq N \}$ is bounded. Also, $\{a_{n}: n < N \}$ is also bounded. The union, $\{a_{n}: n \geq N \} \cup \{a_{n}: n < N \} = \{a_{n}: n \in \bold{N}\}$ is thus bounded. So it has a convergent subsequence $c_n \to L$. For $\varepsilon _1 > 0$ we choose $N \in \bold{N}$ such that $n,m \geq N \implies |a_{n}-a_{m}| < \varepsilon_{1}/2$. So $c_{k} = a_{m_{k}}$ such that $m_{k} \geq N$ and $|c_{m_{k}}-L| < \varepsilon_{1}/2$. For $n \geq N$, we have $|a_{n}-L| = |a_{n}-c_{k}+c_{k}-L| \leq |a_{n}-c_{k}| + |c_{k}-L| < |a_{n}-a_{m_{k}}| + \varepsilon_{1}/2 < \varepsilon_1$. So $a_n \to L$. $\square$

5. That's good. It works.
You say what the $b_n$ in the proof comes from.