# Suppose $0 \leq u_n \leq v_n$ for all $n\in \mathbb{N}$. Prove that if $v_n$ converges to zero, then $u_n$ converges to zero.

So I have an idea of how to do this and I want to make sure I'm right. Here is my attempt:

Proof:

Given: $$v_n$$ converges to zero.

In other words: for all $$\epsilon \in \mathbb{Q}^+$$ there exists a natural number $$N$$, for all naturals $$n$$, such that if $$n\geq N$$ then $$|v_n - 0| < \epsilon$$.

Note that $$0\leq u_n \leq v_n$$ implies $$0\leq |u_n| \leq |v_n|$$, so we get:

$$\epsilon > |v_n - 0| = |v_n| \geq |u_n| = |u_n - 0|$$.

Therefore, for all $$\epsilon \in \mathbb{Q}^+$$ there exists a natural number $$N$$, for all naturals $$n$$, such that if $$n\geq N$$ then $$|u_n - 0| < \epsilon$$.

• Yes, it is correct. Commented Oct 16, 2020 at 7:30
• @xhsbm No problem with the proof, but what is the point of using $\varepsilon \in \mathbb{Q} ^+$? Commented Oct 16, 2020 at 7:32
• @PierreCarre Because we're still constructing the reals. Commented Oct 16, 2020 at 7:35

## 1 Answer

since $$\forall_{n \in \mathbb{N}}\ 0 \le u_n \le v_n => 0 \le lim_{n->\infty} u_n \le lim_{n->\infty} v_n = 0 => lim_{n->\infty} u_n = 0$$