Prove that {$x_n$} is not bounded.

Question

Suppose $\{x_n\}$ is a sequence such that $x_{n+1}-x_n\geq10^{-6}$ for all $n$. Prove that $\{x_n\}$ is not bounded.

This is what I have:

$x_{n+1}-x_n\geq10^{-6}$

$x_{n+1}-x_n\geq\frac{1}{10^{6}}$

$x_{n+1}\geq x_n+\frac{1}{10^{6}}$

$x_n-x_1=x_n-x_{n-1}+x_{n-1}-x_{n-2}+x_{n-2}-x_{n-3}+\cdots\geq\frac{1}{10^{6}} + \frac{1}{10^{6}} +\cdots+\frac{1}{10^{6}}=\frac{n}{10^{6}}$

$x_n\geq x_1 + \frac{1}{10^{6}}$

Since $\frac{n}{10^{6}}$ is an unbounded sequence and $x_n\geq x_1 + \frac{n}{10^{6}}$, Therefore $x_n$ is not bounded.

Is there another way to do this besides the way I did?

Answer 1

This is the good method, but the most "elegant" way to write it is to make an (very easy) induction to prove that the inequality $$x_n \geq x_1 + \frac{n-1}{10^6}$$

holds for every integer $n \in \mathbb{N}^*$.