📂Analysis

Properties of Divergent Real Sequences

Summary¹

Let $\left\{ x_{n} \right\}$, $\left\{ y_{n} \right\}$ be real sequences and let $\lim \limits_{n\to\infty} x_{n}=\infty(-\infty)$. Then the following hold:

• (a) If $\left\{ y_{n} \right\}$ is bounded below (bounded above), then $\lim \limits_{n\to\infty}(x_{n}+y_{n}) = \infty(-\infty)$.

• (b) $\forall \alpha > 0,\quad \lim \limits_{n\to\infty} \alpha x_{n} = \infty (-\infty)$.

• (c) If for every $n\in \mathbb{N}$, there exists a $M_{0} >0$ such that $y_{n} > M_{0}$, then $\lim \limits_{n\to\infty} x_{n}y_{n} = \infty(-\infty)$.

• (d) If $\left\{ y_{n} \right\}$ is bounded and for every $n\in \mathbb{N}$, $x_{n} \ne 0$, then $\lim \limits_{n\to \infty} \dfrac{y_{n}}{x_{n}} = 0$.

Proof

(a)

By assumption, for every $n$, there exists a $M_{0} \in \mathbb{R}$ such that $y_{n} \ge M_{0}$. Fix any $M \in \mathbb{R}$ and let $M_{1}=M-M_{0}$. Since $\left\{ x_{n} \right\}$ diverges, there exists a $N \in \mathbb{N}$ such that the following holds:

$$n \ge N \implies x_{n} > M_{1}$$

Thus, the following equation holds, and the proof is complete.

$$n \ge N \implies x_{n} + y_{n}> M_{1} + M_{0} > M$$

■

(b)

Let $M\in \mathbb{R}$ and for any $\alpha > 0$, let $M_{1}= M / \alpha$. Since $\left\{ x_{n} \right\}$ diverges, there exists a $N \in \mathbb{N}$ such that the following holds:

$$n \ge N \implies x_{n} > M_{1}$$

Thus, the following equation holds, and the proof is complete.

$$n \ge N \implies \alpha x_{n} > \alpha M_{1} = M$$

■

(c)

Let $M\in \mathbb{R}$ and $M_{1}= M / M_{0}$. Since $\left\{ x_{n} \right\}$ diverges, there exists a $N \in \mathbb{N}$ such that the following holds:

$$n \ge N \implies x_{n} > M_{1}$$

Thus, the following equation holds, and the proof is complete.

$$n \ge N \implies x_{n}y_{n} > M_{1}M_{0} = M$$

■

(d)

Let $\epsilon > 0$. Since $\left\{ y_{n} \right\}$ is bounded, there exists a $M_{0}>0$ such that $| y_{n} | \le M_{0}$. Choose a sufficiently large $M_{1}>0$ satisfying $\dfrac{M_{0}}{M_{1}} < \epsilon$. Since $\left\{ x_{n} \right\}$ diverges, there exists a $N \in \mathbb{N}$ such that the following holds:

$$n \ge N \implies x_{n} > M_{1}$$

Thus, the following equation holds, and the proof is complete.

$$n \ge N \implies \left| \dfrac{y_{n}}{x_{n}} - 0 \right| = \dfrac{| y_{n} |}{x_{n}} < \dfrac{M_{0}}{M_{1}} < \epsilon$$

■