The Cauchy Criterion
Up until now, we've needed to know the limit value $\ell$ in order to verify that a sequence converges to $\ell$. But what if we don't know what the limit is? The Cauchy Criterion provides a way to determine whether a sequence converges without knowing the limit in advance.
Intuition: A sequence converges if its terms "cluster together" as we move further along the sequence. Specifically, if the sequence converges, then terms far out must be close to each other.
Definition: Cauchy Sequence
A sequence $(a_n)$ is a Cauchy sequence if for every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that whenever $m, n \geq N$, we have
$$$$Theorem 1: Convergent $\Rightarrow$ Cauchy
If $(a_n) \to a$, then $(a_n)$ is a Cauchy sequence.
Proof
Suppose $(a_n) \to a$. Let $\varepsilon > 0$.
Step 1: Apply convergence definition
Since $(a_n) \to a$, there exists $N \in \mathbb{N}$ such that whenever $n \geq N$, we have
$$|a_n - a| < $$Step 2: Use triangle inequality
For $m, n \geq N$:
$$|a_m - a_n| = |a_m - a + a - a_n| \leq |a_m - a| + |a_n - a|$$Step 3: Estimate the bound
Therefore, $(a_n)$ is a Cauchy sequence. $\square$
Theorem 2 (The Cauchy Criterion): Cauchy $\Rightarrow$ Convergent
A sequence $(a_n)$ is Cauchy if and only if $(a_n)$ converges.
Why This Matters
Theorem 1 + Theorem 2 together tell us that a sequence converges to some limit $\ell$ if and only if its terms cluster together. We don't need to know $\ell$ in advance.
Lemma: Every Cauchy Sequence is Bounded
Proof (Fill in the steps)
Suppose $(a_n)$ is Cauchy. Choose $\varepsilon = 1$. There exists $N$ such that for $m, n \geq N$:
$$|a_m - a_n| < 1$$Step 1: Bound terms after $N$
For $n \geq N$, all such terms satisfy $|a_n - a_N| < 1$, so
$$|a_n| < |a_N| + 1$$Step 2: Bound all terms
Define $M = \max\{|a_1|, |a_2|, \ldots, |a_{N-1}|, |a_N| + 1\}$. Then
$$|a_n| \leq M \quad \text{for all } n \in \mathbb{N}$$Therefore, $(a_n)$ is bounded. $\square$
Why the Cauchy Criterion Works in $\mathbb{R}$
The proof that Cauchy $\Rightarrow$ Convergent relies on the Bolzano-Weierstrass Theorem: every bounded sequence has a convergent subsequence.
Theorem: Bolzano-Weierstrass (Version for Bounded Sequences)
Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.
Proof Outline: Why Cauchy $\Rightarrow$ Convergent (using BW)
Given: $(a_n)$ is Cauchy.
Step 1: $(a_n)$ is bounded
Use the Lemma above.
Step 2: Extract a convergent subsequence
By Bolzano-Weierstrass, $(a_n)$ has a subsequence $(a_{n_k}) \to L$ for some $L \in \mathbb{R}$.
Step 3: Show the full sequence converges to $L$
Let $\varepsilon > 0$. Since $(a_n)$ is Cauchy, there exists $N_1$ such that
$$|a_m - a_n| < \frac{\varepsilon}{2} \quad \text{for all } m,n \geq N_1$$Since $(a_{n_k}) \to L$, there exists $K$ such that
$$|a_{n_K} - L| < $$For $n \geq \max\{N_1, n_K\}$:
Therefore, $(a_n) \to L$. $\square$
Example 1: A Cauchy Sequence
Show that $a_n = \dfrac{1}{n}$ is Cauchy.
Solution:
Let $\varepsilon > 0$. We want $|a_m - a_n| < \varepsilon$ for $m, n \geq N$.
For $m > n$:
$$|a_m - a_n| = \left|\frac{1}{m} - \frac{1}{n}\right| = \frac{1}{n} - \frac{1}{m} < \frac{1}{n}$$Choose $N$ such that . Then for $m, n \geq N$:
$$|a_m - a_n| < $$Therefore, $(a_n)$ is Cauchy.
Example 2: A Non-Cauchy Sequence
Show that $b_n = (-1)^n$ is not Cauchy.
Solution:
To show $(b_n)$ is not Cauchy, find $\varepsilon > 0$ such that large $m, n$, we have $|b_m - b_n| \geq \varepsilon$.
Choose $\varepsilon = 1$. Take $m = 1$ (so $b_m = -1$) and $n = 2$ (so $b_n = 1$). Then:
$$|b_m - b_n| = |-1 - 1| = 2 > 1 = \varepsilon$$In fact, $|b_m - b_n| = 2$ for all $m$ odd and $n$ even. Therefore, $(b_n)$ is not Cauchy.
Exercise 1
Show that the sequence $a_n = \dfrac{n}{n+1}$ is Cauchy.
Exercise 2
Determine whether the sequence $b_n = \sqrt{n}$ is Cauchy. Justify your answer.
The Cauchy Criterion provides an intrinsic condition for convergence: A sequence converges if and only if its terms eventually cluster together, without needing knowledge of the limit value itself.
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