> [!definition] > > Let $(x_n)$ be a [[Sequence|sequence]]. It is *properly* divergent if > $ > \forall \alpha \in \real, \exists N \in \nat: \forall n \ge N, x_n \ge \alpha > $ > or > $ > \forall \alpha \in \real, \exists N \in \nat: \forall n \ge N, x_n \le \alpha > $ > This is denoted as > $ > \limv{n}x_n = \infty \quad \limv{n}x_n = -\infty > $ > [!theorem] > > Let $(x_n)$ be a sequence. It can only be *properly* divergent in a given direction if it is not bounded in that direction. If $(x_n)$ is monotone and unbounded, then it is properly divergent. > [!theorem] > > Let $(x_n)$ and $(y_n)$ be two sequences with $x_n \ge y_n \forall n \in \nat$. Then if $\limv{n}y_n = \infty$, $\limv{n}x_n = \infty$, and if $\limv{n}x_n = - \infty$, $\limv{n}y_n = -\infty$. > [!theorem] > > Let $(x_n)$ be a properly diverging sequence with $\limv{n}x_n = \infty$, and $c \in \real$, $c \ne 0$. Then, $\limv{n}cx_n = \infty$ if $c > 0$ and $\limv{n}cx_n = -\infty$ if $c < 0$. > [!theorem] > > Let $(x_n)$ and $(y_n)$ be increasing, strictly positive sequences. Suppose that $\exists\limv{n}\frac{x_n}{y_n} = L > 0$. Then > $ > \limv{n}x_n = \infty \Leftrightarrow \limv{n}y_n = \infty > $ > > *Proof*. Take $\varepsilon = L/2$. Then > $ > \exists N \in \nat: \forall n \ge N, L - L/2 < \frac{x_n}{y_n} < L + L/2 > $ > or > $ > \frac{L}{2}y_n < x_n < \frac{3L}{2}y_n > $ > therefore $\limv{n}x_n = \infty \Leftrightarrow \limv{n}y_n = \infty$, enforced by this inequality. > [!theorem] > > If $\limv{n}\frac{x_n}{y_n} = 0$ and $\limv{n}x_n = \infty$, then $\limv{n}y_n = \infty$. > > *Proof*. Take $\varepsilon = 1$, then > $ > \exists N \in \nat: \forall n \ge N, \abs{\frac{x_n}{y_n}} < 1 \Leftrightarrow \abs{x_n} < \abs{y_n} > $ > which conducts the proper divergence. > [!theorem] > > Let $(x_n)$ be a properly divergent sequence and $(y_n)$ be a bounded sequence. Then $\limv{n}(x_n + y_n) = \pm \infty$.