> [!theorem] Monotone Convergence Theorem > > If a [[Sequence|sequence]] is monotone and bounded, then it [[Limit|converges]] to its [[Supremum and Infimum|limit point]] in that direction. > > *Proof*. Suppose that $(x_n)$ is decreasing and bounded from below. Let $A = \{x_n: n \in \nat^+\}$ be the set of all elements in the sequence. By completeness, $x = \inf A$ exists. > > By definition of the lower bound, $x \le a \forall a \in A$. Since $x$ is also the infimum, > $ > \forall \varepsilon > 0, \exists x_N \in A: x_N < x + \varepsilon > $ > As the sequence is decreasing > $ > \forall n \ge N, x_n \le x_{N} > $ > Therefore, > $ > \forall \varepsilon > 0, \exists N \in \nat: \forall n \ge N, |x_n - x| < \varepsilon > $ > [!theorem] > > Let $(x_n)$ be a sequence. $(x_n)$ is *eventually* increasing/decreasing if after a certain threshold $t \in \nat$, it is increasing or decreasing ($x_{n + 1} \ge x_{n}$/$x_{n + 1} \le x_{n}$ $\forall n \ge t$). Let $Y = \{x_n: n \ge t\}$ be the set of elements in the sequence past that threshold. If $(x_n)$ is eventually increasing/decreasing, and that $Y$ is bounded from above/below, then $\limv{n}x_n = \sup Y$/$\limv{n}x_n = \inf Y$. > > *Proof*. Considered a shifted sequence $(y_n) = (x_{n + t})$ with $y_n = x_{n + t}$. Then, $(y_n)$ is increasing/decreasing for all $n \in \nat$. As $Y$ is bounded from above/below, $(y_n)$ is also bounded from above/below. Therefore, by the monotone convergence theorem, $\limv{n}y_n = \sup X$/$\limv{n}y_n = \inf X$. > > Since $(y_n)$ is just a shifted version of $(x_n)$, $\limv{n}y_n = \limv{n}x_n$, which would be $\sup Y$ or $\inf Y$. > > Let $X = \{x_n: n \in \nat\}$. Note that $\lim_{x \to \infty}x_n$ is not necessarily $\sup X$ or $\inf X$. Suppose that $(x_n)$ is bounded above and eventually increasing with $t > 1$, and that $x_1 \gt x_{n} + 1 \forall n \gt 1$, then $x_1 = \sup X$ because it's the maximum element. Let $\varepsilon = \frac{1}{2}$, then since $x_1 - x_n > 1 \forall n > 1$, there exists no $N: \forall n \ge N, |x_n - x_1| < \varepsilon$.