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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?

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  • $\begingroup$ 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 $n$. $\endgroup$
    – YiFan Tey
    Commented Dec 26, 2020 at 10:11

1 Answer 1

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Your proof is correct, but perhaps you want to rewrie the statement as such:

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 $|{-}3|>|{-}1|$).

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