cauchy sequence

Cauchy sequence

A sequence $(a_n{)}_{n\in \mathbb{N}}\subset \mathbb{C}$ is called a Cauchy sequence if

$$\forall ϵ>0\exists n_0\in \mathbb{N}\forall n,m\ge n_0:|a_n-a_m|<ϵ$$

That is, the terms of a Cauchy sequence are pairwise close to each other for large $n$.

Compare with convergence: convergence says terms get close to a specific value $a$. Cauchy says terms get close to each other. The power of the Cauchy criterion is that you can prove convergence without knowing the limit:

If $|a_{n+1}-a_n|\le C\cdot |a_n-a_{n-1}|$ for some $0<C<1$

Each gap is at most a fixed fraction of the previous gap. By induction: $|a_{n+1}-a_n|\le C^{n-1}|a_2-a_1|$. Since $C<1$, the gaps shrink to zero geometrically, so the terms bunch up: the sequence is Cauchy, hence convergent.

This works for any such sequence. You never need to know the limit.

Cauchy criterion

$$(a_n)\text{ is convergent}⟺(a_n)\text{ is a Cauchy sequence}$$

Proof sketch

$(\Rightarrow )$ If $a_n\to a$, then for large $n,m$:

$$|a_n-a_m|\le |a_n-a|+|a-a_m|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon$$

$(\Leftarrow )$ Cauchy $⟹$ bounded (fix $\epsilon =1$, then all terms past some $n_0$ are within $1$ of $a_{n_0}$). Bounded $⟹$ has a convergent subsequence by Bolzano-Weierstrass. Call its limit $a$. Then:

$$|a_n-a|\le |a_n-a_{n_k}|+|a_{n_k}-a|$$

Both terms $<\epsilon /2$ for large enough $n$ and $n_k$ (first by Cauchy, second by subsequence convergence).

Neighbors getting close is not enough

$a_n=\sqrt{n}$: consecutive terms satisfy $|a_{n+1}-a_n|=\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$ (by the conjugate trick), but the sequence diverges.

Cauchy requires all pairs $|a_n-a_m|<\epsilon$ for $n,m\ge n_0$, not just neighbors. Here $|a_{4n}-a_n|=\sqrt{4n}-\sqrt{n}=\sqrt{n}\to \infty$.

$a_n=\frac{1}{n}$ is Cauchy

For $m\ge n$: $|a_n-a_m|=\frac{1}{n}-\frac{1}{m}\le \frac{1}{n}$.

Given $\epsilon >0$, pick $n_0>\frac{1}{\epsilon }$. Then for all $n,m\ge n_0$:

$$|a_n-a_m|\le \frac{1}{n}\le \frac{1}{n_0}<\epsilon$$

Completeness

The Cauchy criterion relies on $\mathbb{R}$ being complete. In $\mathbb{Q}$, a Cauchy sequence of rationals can “converge” to an irrational that isn’t in $\mathbb{Q}$, so Cauchy $\nRightarrow$ convergent there.
This is gives us an equivalent way to define completeness: a space is complete iff every Cauchy sequence converges.

Applying the cauchy criterion to partial sums.

A series is convergent iff the sequence of partial sums is a cauchy sequence.
Let ${\sum }_{k=1}^{\infty }{a}_k$ be a series. then we have that ${\sum }_{k=1}^{\infty }{a}_k$ is convergent iff

$$\begin{array}{rl}\forall ϵ>0\exists n_0\in \mathbb{N}\forall m>n\ge n_0:& |\sum _{k=n+1}^m{a}_k|<ϵ\\ & =|\sum _{k=1}^m{a}_k-\sum _{k=1}^n{a}_k|<ϵ\\ & =|s_m-s_n|<ϵ\end{array}$$

State the Cauchy condition for a sequence $(a_n)$ in quantifiers.

$$\forall ϵ>0\exists n_0\in \mathbb{N}\forall n,m\ge n_0:|a_n-a_m|<ϵ$$

In words, what does it mean for a sequence to be Cauchy? eachother for large $n$.

Terms are pairwise close to

What can you do with the Cauchy criterion that the definition of convergence does not let you do?

Prove convergence without knowing the limit.

Convergence and Cauchy both say "the terms get close." Close to what, in each case?

Convergence: close to a specific value, the limit
Cauchy: close to eachother

A sequence has $|a_{n+1}-a_n|\to 0$. Is it necessarily Cauchy? Give the proof / counterexample.

Circular transclusion detected: general/cauchy-sequence

What is the Cauchy criterion for series?

A series ${\sum }_{k=1}^{\infty }{a}_k$ is convergent iff

$$\begin{array}{rl}\forall ϵ>0\exists n_0\in \mathbb{N}\forall m>n\ge n_0:& |\sum _{k=n+1}^m{a}_k|<ϵ\\ & =|\sum _{k=1}^m{a}_k-\sum _{k=1}^n{a}_k|<ϵ\\ & =|s_m-s_n|<ϵ\end{array}$$