> [!NOTE] **Theorem** (Monotone bounded sequences converge) > Any [[Monotonic Sequence of Real Numbers|monotonic sequence]] $(a_{n})_{n=m}^{\infty}$ is [[Bounded Sequence|bounded]] iff it converges with $\lim_{ n \to \infty } a_{n} = \sup (a_{n})_{n=m}^{\infty}$taking the [[Suprema and Infima of Sequences|supremum of the sequence]]. > *Proof*. ($\implies$) Consider the set $A = \{a_{n}: n \in \mathbb{N}\}$ which is non-empty and bounded above (by assumption), so by the ([[Real numbers|LUBA]]), it has a finite supremum $l := \sup A$. > > Since $l$ is an upper bound for all elements of A,$a_{n}\leq l \quad \text{for every } n \in \mathbb{N} \quad \tag{1}$ Now take $\varepsilon>0$. Since $l$ is the supremum, there exists an element $a$ of $A$ such that $a > l-\epsilon,$ i.e. there exists $N$ such that $a_{N} > l-\epsilon$Since $(a_{n})$ is increasing, we know that if $n\geq N$ then $a_{n}\geq a_{N}>l-\epsilon$Combining this with $(1)$ it follows that for all $n\geq N,$ we have $|a_{n}-l| <\varepsilon$, which shows that $a_{n} \to l$ as $n \to \infty.$ > > ($\Longleftarrow$) Choose $R \in \mathbb{R}$. Since $(a_{n})$ is not bounded above, $\exists N$ such that $a_{N}>R$. > Now since $(a_{n})$ is increasing, it follows that. $a_{n}>R$ for all $n\geq N$; so $a_{n}\to \infty$. > It follows that if $(a_{n})$ is a decreasing sequence then it converges $\iff$ it's bounded below. > # Applications - [[Series with Non-Negative Terms Converges Iff Partial Sums Are Bounded Above]]. - [[Bolzano-Weierstrass Theorem (Sequential Compactness of The Reals)]] - [[Comparison Test for Series With Non-Negative Terms]] - [[Alternating series test]].