[proofplan] For an increasing bounded [sequence](/page/Sequence), the least upper bound axiom produces the only possible limit, namely the [supremum](/page/Supremum%20and%20Infimum) of its range. The definition of supremum gives a sequence term within any $\varepsilon$ of that supremum from below, and monotonicity keeps all later terms in the same interval. The decreasing case is reduced to the increasing case by negating the sequence, which converts lower bounds into upper bounds and infima into negatives of suprema. [/proofplan] [step:Choose the supremum as the candidate limit in the increasing case] Assume that $(a_n)_{n=1}^{\infty}$ is increasing and bounded above. Define the set \begin{align*} A := \{a_n : n \in \mathbb{N}\} \subset \mathbb{R}. \end{align*} The set $A$ is non-empty because $a_1 \in A$, and $A$ is bounded above by the boundedness hypothesis. By the [least upper bound axiom](/page/Least%20Upper%20Bound%20Axiom), there exists $l \in \mathbb{R}$ such that \begin{align*} l = \sup A = \sup\{a_n : n \in \mathbb{N}\}. \end{align*} [guided] Assume that $(a_n)_{n=1}^{\infty}$ is increasing and bounded above. To prove [convergence](/page/Convergence%20%28Real%20Sequences%29), we first identify the only reasonable candidate for the limit. Define \begin{align*} A := \{a_n : n \in \mathbb{N}\} \subset \mathbb{R}. \end{align*} This notation records the range of the sequence as a subset of $\mathbb{R}$. The set $A$ is non-empty because it contains $a_1$. It is bounded above because the sequence is bounded above by hypothesis, meaning that there exists $M \in \mathbb{R}$ such that $a_n \le M$ for every $n \in \mathbb{N}$. The least upper bound axiom applies to every non-empty subset of $\mathbb{R}$ that is bounded above. Therefore there exists a real number $l \in \mathbb{R}$ satisfying \begin{align*} l = \sup A = \sup\{a_n : n \in \mathbb{N}\}. \end{align*} This number is the candidate limit: it is an upper bound for every term of the sequence, and by the defining property of the supremum no smaller number is an upper bound. [/guided] [/step] [step:Use the supremum property to find a term within $\varepsilon$ from below] Let $\varepsilon > 0$. Since $l = \sup A$, the number $l - \varepsilon$ is not an upper bound for $A$. Hence there exists $N \in \mathbb{N}$ such that \begin{align*} a_N > l - \varepsilon. \end{align*} [guided] Let $\varepsilon > 0$. We want to show that all sufficiently late terms of the sequence lie within distance $\varepsilon$ of $l$. The supremum property gives the first such term. Since $\varepsilon > 0$, we have $l - \varepsilon < l$. If $l - \varepsilon$ were an upper bound for $A$, then $l$ would not be the least upper bound of $A$, because a strictly smaller upper bound would exist. Therefore $l - \varepsilon$ is not an upper bound for $A$. By the definition of “not an upper bound,” there is an element of $A$ strictly larger than $l - \varepsilon$. Since every element of $A$ has the form $a_n$ for some $n \in \mathbb{N}$, there exists $N \in \mathbb{N}$ such that \begin{align*} a_N > l - \varepsilon. \end{align*} This index $N$ is the point after which monotonicity will keep the sequence above $l - \varepsilon$. [/guided] [/step] [step:Use monotonicity to keep all later terms between $l - \varepsilon$ and $l$] For every $n \ge N$, monotonicity gives $a_n \ge a_N$, and therefore \begin{align*} a_n > l - \varepsilon. \end{align*} Also, since $a_n \in A$ and $l$ is an upper bound for $A$, we have $a_n \le l$. Thus, for every $n \ge N$, \begin{align*} l - \varepsilon < a_n \le l. \end{align*} Consequently, \begin{align*} |a_n - l| < \varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, $a_n \to l$ as $n \to \infty$ by the definition of convergence of a sequence. [guided] Fix $n \ge N$. Since the sequence is increasing, the inequality $n \ge N$ implies \begin{align*} a_n \ge a_N. \end{align*} From the previous step, $a_N > l - \varepsilon$, so we obtain \begin{align*} a_n > l - \varepsilon. \end{align*} This gives the lower side of the desired estimate. For the upper side, use the fact that $l = \sup A$. In particular, $l$ is an upper bound for $A$. Since $a_n \in A$, it follows that \begin{align*} a_n \le l. \end{align*} Combining the two inequalities gives, for every $n \ge N$, \begin{align*} l - \varepsilon < a_n \le l. \end{align*} Subtracting $l$ throughout places $a_n - l$ between $-\varepsilon$ and $0$, hence \begin{align*} |a_n - l| < \varepsilon. \end{align*} This is exactly the $\varepsilon$-definition of convergence for the sequence $(a_n)_{n=1}^{\infty}$. Since the argument works for every $\varepsilon > 0$, we conclude that \begin{align*} a_n \to l = \sup\{a_n : n \in \mathbb{N}\} \end{align*} as $n \to \infty$. [/guided] [/step] [step:Reduce the decreasing case to the increasing case by negating the sequence] Assume that $(a_n)_{n=1}^{\infty}$ is decreasing and bounded below. Define a sequence $(b_n)_{n=1}^{\infty}$ by \begin{align*} b_n := -a_n \end{align*} for every $n \in \mathbb{N}$. Since $a_{n+1} \le a_n$, we have $b_{n+1} = -a_{n+1} \ge -a_n = b_n$, so $(b_n)_{n=1}^{\infty}$ is increasing. If $m \in \mathbb{R}$ is a lower bound for $\{a_n : n \in \mathbb{N}\}$, then $a_n \ge m$ for every $n \in \mathbb{N}$, hence $b_n \le -m$ for every $n \in \mathbb{N}$. Thus $(b_n)_{n=1}^{\infty}$ is bounded above. By the increasing case already proved, \begin{align*} b_n \to \sup\{b_n : n \in \mathbb{N}\}. \end{align*} Since $b_n = -a_n$, this is equivalent to \begin{align*} a_n \to -\sup\{-a_n : n \in \mathbb{N}\}. \end{align*} Finally, \begin{align*} -\sup\{-a_n : n \in \mathbb{N}\} = \inf\{a_n : n \in \mathbb{N}\}, \end{align*} because negation reverses order. Therefore every decreasing sequence bounded below converges to its infimum. [guided] Assume that $(a_n)_{n=1}^{\infty}$ is decreasing and bounded below. We convert this case into the increasing case by changing signs. Define the sequence $(b_n)_{n=1}^{\infty}$ by \begin{align*} b_n := -a_n \end{align*} for every $n \in \mathbb{N}$. We verify the hypotheses of the increasing case for $(b_n)_{n=1}^{\infty}$. Since $(a_n)_{n=1}^{\infty}$ is decreasing, we have $a_{n+1} \le a_n$ for every $n \in \mathbb{N}$. Multiplying this inequality by $-1$ reverses the order, so \begin{align*} b_{n+1} = -a_{n+1} \ge -a_n = b_n. \end{align*} Thus $(b_n)_{n=1}^{\infty}$ is increasing. Next, because $(a_n)_{n=1}^{\infty}$ is bounded below, there exists $m \in \mathbb{R}$ such that $m \le a_n$ for every $n \in \mathbb{N}$. Multiplying by $-1$ gives \begin{align*} b_n = -a_n \le -m \end{align*} for every $n \in \mathbb{N}$. Hence $(b_n)_{n=1}^{\infty}$ is bounded above. The increasing case applies, so \begin{align*} b_n \to \sup\{b_n : n \in \mathbb{N}\}. \end{align*} Substituting $b_n = -a_n$, this says \begin{align*} -a_n \to \sup\{-a_n : n \in \mathbb{N}\}. \end{align*} Multiplying the convergent sequence by $-1$ gives \begin{align*} a_n \to -\sup\{-a_n : n \in \mathbb{N}\}. \end{align*} It remains to identify this number with the infimum of the original sequence. Let \begin{align*} \alpha := \inf\{a_n : n \in \mathbb{N}\}. \end{align*} A real number $r$ is a lower bound for $\{a_n : n \in \mathbb{N}\}$ exactly when $r \le a_n$ for every $n \in \mathbb{N}$, which is equivalent after multiplying by $-1$ to $-r \ge -a_n$ for every $n \in \mathbb{N}$. Thus lower bounds of $\{a_n : n \in \mathbb{N}\}$ correspond under negation to upper bounds of $\{-a_n : n \in \mathbb{N}\}$. Therefore \begin{align*} -\sup\{-a_n : n \in \mathbb{N}\} = \inf\{a_n : n \in \mathbb{N}\}. \end{align*} Consequently the decreasing sequence converges to its infimum. [/guided] [/step]