The Cauchy Criterion for Convergence

Recall that a sequence of real numbers $(a_{n})$ is said to be a Cauchy Sequence if $\forall \epsilon >0$ there exists an $N\in \mathbb {N}$ such that if $m,n\geq N$ then $\mid a_{n}-a_{m}\mid <\epsilon$. We already noted that every convergent sequence of real numbers is Cauchy, and that every Cauchy sequence of real numbers is bounded. We will now look at another important theorem known as the Cauchy Convergence Criterion.

Theorem (Cauchy Convergence Criterion): If $(a_{n})$ is a sequence of real numbers, then $(a_{n})$ is convergent if and only if $(a_{n})$ is a Cauchy sequence.

Note that the Cauchy Convergence Criterion will allow us to determine whether a sequence of real numbers is convergent whether or not we have a suspected limit in mind for a sequence.

• Proof: $\Rightarrow$ Suppose that $(a_{n})$ is a convergent sequence of real numbers.

• $\Leftarrow$ Suppose that $(a_{n})$ is a Cauchy sequence. We want to show that $(a_{n})$ is thus convergent to some real number in $\mathbb {R}$. Now since $(a_{n})$ is a Cauchy sequence it follows that $(a_{n})$ is bounded. Since $(a_{n})$ is bounded, it follows from <a href="/the-bolzano-weierstrass-theorem">The Bolzano-Weierstrass Theorem</a> that there exists a subsequence of $(a_{n})$, call it $(a_{n_{k}})$ that converges to some real number $A^{*}$.

• Since $(a_{n})$ is a Cauchy sequence, then $\forall \epsilon >0$ there exists an $N\in \mathbb {N}$ such that if $m,n\geq N$ then $\mid a_{n}-a_{m}\mid <\epsilon$. Choose $\epsilon _{1}={\frac {\epsilon }{2}}>0$, and so there exists an $N_{1}\in \mathbb {N}$ such that if $m,n\geq N_{1}$ then $\mid a_{n}-a_{m}\mid <\epsilon _{1}={\frac {\epsilon }{2}}$.

• Now look at the subsequence $(a_{n_{k}})$ which converges to $A^{*}$. There thus exists a natural number $K\geq N_{1}$ where $K$ belongs to the set of indices $\{n_{1},n_{2},...\}$ such that $\mid a_{K}-A^{*}\mid <{\frac {\epsilon }{2}}$.

• Since $K\geq N_{1}$ then if we substitute $m=K$ we have that for $n\geq N_{1}$:

${\begin{aligned}\mid a_{n}-a_{K}\mid <{\frac {\epsilon }{2}}\end{aligned}}$

• And so for $n\geq N_{1}$ we have that:

${\begin{aligned}\quad \mid a_{n}-A^{*}\mid =\mid a_{n}-a_{K}+a_{K}-A^{*}\mid \leq \mid a_{n}-a_{K}\mid +\mid a_{K}-A^{*}\mid <{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

• Therefore $\lim _{n\to \infty }a_{n}=A^{*}$, and so $(a_{n})$ is convergent to the real number $A^{*}$. $\blacksquare$

We will summarize the lemma from the Cauchy Sequences and the Cauchy Convergent Criterion as follows:

• Any sequence of real numbers $(a_{n})$ that is convergent is also a Cauchy sequence.

• Any Cauchy sequence of real numbers $(a_{n})$ is also a convergent sequence.

Licensing

Content obtained and/or adapted from:

• The Cauchy Convergence Criterion, mathonline.wikidot.com under a CC BY-SA license