# Proof of theorem about sequence. Check my proof.

Prove this theorem:

Let $$s_n, t_n$$ be sequences on real number. If there exist $$N\in \mathbb N$$ such that $$0\leq s_n\leq t_n$$ and $$\lim \limits_{n\to \infty} t_n=0$$ for all $$n\geq N$$, then $$\lim \limits_{n\to \infty} s_n=0$$.

I have tried as below.

$$\lim \limits_{n\to \infty} t_n=0$$ means for all $$\varepsilon>0$$ there exist $$N\in \mathbb N$$ such that $$\vert t_n\vert<\varepsilon$$

for all $$n\geq N$$.

Since $$s_n\leq t_n$$, we have $$\vert s_n\vert \leq \vert t_n\vert$$. So, $$\vert s_n-0\vert=\vert s_n\vert \leq \vert t_n\vert<\varepsilon.$$ Thus, $$\lim \limits_{n\to \infty} s_n=0.$$

I'm not sure with my answer. Anyone can check my proof? Is it correct proof?

• The phrasing "$\lim_{n\to\infty}t_n=0$ for all $n\geq N$" sounds really weird, as the limit is a property of the sequence independent of $n$. Dec 26, 2020 at 10:11

Let $$s_n,t_n$$ be sequences of real numbers. Suppose that $$\lim_{n\to\infty}t_n=0$$ and there exists $$N\in\mathbb N$$ so that $$0\leq s_n\leq t_n$$ for all $$n>N$$. Then $$\lim_{n\to\infty}s_n=0$$.
The point is that having limiting value $$0$$ is a property of the sequence $$(t_n)$$ independent of $$n$$, so your original phrasing sounds awkward. Another minor criticism I would have is that when you reiterated the definition of $$\lim t_n=0$$, you implicitly assumed the $$N$$ in the definition is the same number as the $$N$$ given to you in the question, which is not necessarily always true---but in this case your proof is not affected. You also want to make explicit that the logical implication $$s_n\leq t_n\implies |s_n|\leq |t_n|$$ is dependent on the fact that $$s_n,t_n\geq 0$$, because otherwise this is not true (e.g. $$-3<-1$$ but $$|{-}3|>|{-}1|$$).