[proofplan] We prove both directions of the equivalence between convergence and the Cauchy property for real sequences. The forward direction is a direct $\epsilon/2$-argument using the triangle inequality. For the reverse, we first show Cauchy sequences are bounded, then apply the [Bolzano-Weierstrass Theorem](/theorems/171) to extract a convergent subsequence, and finally bootstrap from subsequential convergence to full convergence using the Cauchy property. [/proofplan] [step:Show that convergent sequences are Cauchy via the triangle inequality] Assume $a_n \to a$ for some $a \in \mathbb{R}$. Let $\epsilon > 0$ be given. There exists $N \in \mathbb{N}$ such that $|a_n - a| < \epsilon/2$ for all $n \geq N$. For any $n, m \geq N$, the triangle inequality gives \begin{align*} |a_n - a_m| \leq |a_n - a| + |a - a_m| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} Therefore $(a_n)$ is a [Cauchy sequence](/pages/1201). [/step] [step:Prove that Cauchy sequences are bounded] Assume $(a_n)$ is a [Cauchy sequence](/pages/1201). Taking $\epsilon = 1$ in the Cauchy definition, there exists $N \in \mathbb{N}$ such that $|a_n - a_m| < 1$ for all $n, m \geq N$. Fixing $m = N$, the triangle inequality gives $|a_n| \leq |a_n - a_N| + |a_N| < 1 + |a_N|$ for all $n \geq N$. Define \begin{align*} K = \max\{|a_1|, |a_2|, \ldots, |a_{N-1}|, 1 + |a_N|\}. \end{align*} Then $|a_n| \leq K$ for all $n \in \mathbb{N}$, so $(a_n)$ is bounded. [guided] A [Cauchy sequence](/pages/1201) has terms that are eventually close to each other, but we need to confirm they do not escape to infinity. The idea is: fix one term $a_N$ and use the Cauchy property to bound everything after it, then handle the finitely many initial terms separately. Taking $\epsilon = 1$, there exists $N \in \mathbb{N}$ with $|a_n - a_m| < 1$ for all $n, m \geq N$. Setting $m = N$: \begin{align*} |a_n| \leq |a_n - a_N| + |a_N| < 1 + |a_N| \quad \text{for all } n \geq N. \end{align*} The terms $a_1, \ldots, a_{N-1}$ are finitely many, so we take the maximum: \begin{align*} K = \max\{|a_1|, |a_2|, \ldots, |a_{N-1}|, 1 + |a_N|\}. \end{align*} Then $|a_n| \leq K$ for all $n$. This "finitely many exceptions" trick is standard: the Cauchy condition controls the tail, and the finite initial segment is bounded by fiat. [/guided] [/step] [step:Bootstrap from subsequential convergence to full convergence] By the previous step, $(a_n)$ is bounded. By the [Bolzano-Weierstrass Theorem](/theorems/171), there exists a convergent subsequence $(a_{n_j})_{j=1}^{\infty}$ with [limit](/pages/1146) $a \in \mathbb{R}$. Let $\epsilon > 0$ be given. Since $(a_n)$ is Cauchy, there exists $N_1 \in \mathbb{N}$ such that $|a_n - a_m| < \epsilon/2$ for all $n, m \geq N_1$. Since $a_{n_j} \to a$, there exists $J \in \mathbb{N}$ such that $n_J \geq N_1$ and $|a_{n_j} - a| < \epsilon/2$ for all $j \geq J$. For any $n \geq N_1$, the triangle inequality gives \begin{align*} |a_n - a| \leq |a_n - a_{n_J}| + |a_{n_J} - a| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} Therefore $a_n \to a$. [guided] We know $(a_n)$ is bounded and Cauchy. The [Bolzano-Weierstrass Theorem](/theorems/171) extracts a convergent subsequence $(a_{n_j}) \to a$. The question is: does the full [sequence](/pages/1149) also converge to $a$? The Cauchy property is exactly what bridges the gap. For any $n$ far enough out, we can find a subsequential term $a_{n_J}$ that is close to both $a_n$ (by the Cauchy property) and $a$ (by subsequential convergence). Formally: Choose $N_1$ so that $|a_n - a_m| < \epsilon/2$ for all $n, m \geq N_1$. Choose $J$ so that $n_J \geq N_1$ and $|a_{n_J} - a| < \epsilon/2$. Then for $n \geq N_1$, both $n$ and $n_J$ exceed $N_1$, so: \begin{align*} |a_n - a| \leq |a_n - a_{n_J}| + |a_{n_J} - a| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} The subsequential limit acts as a "relay point" -- the Cauchy condition gets us from $a_n$ to $a_{n_J}$, and subsequential convergence gets us from $a_{n_J}$ to $a$. [/guided] [/step]