proving limits of sequences using cauchy criterion

Question

The question is as follows:

Let $\{s_n\}^∞_1$ be a sequence of real numbers such that $s_n > 0$ for $n > k$. Suppose $s = lim_{n→∞}s_n$ exists and is a finite real number. Prove that $s ≥ 0$.

I have an intuition that this is probably most easily solved through contradiction. Given that for $n>k$, $s_n>0$ then by the axiom of archimedes $|s_n|>1/N$ for all $N\epsilon\mathbb{N}$ so it seems to follow that since all $s_n$ are eventually $|s_n|>1/N$ it would be impossible for the cauchy criterion to hold with an $s_n<0$ such that $|s_n-s_m|<1/n$

How does that sound? Could someone help me formalize this, or give some help for a better solution? Thanks!

Answer 1

Suppose on the contrary that $s<0$, then we have $N>0$ such that $n>N$, then $|s_n -s| < |s|$, $$s-|s|<s_n <s+|s|$$

$$2s<s_n <0$$ Having $$s_n < 0$$ is a contradiction.

Answer 2

If possible let $s<0$ then choose $\epsilon =\frac{|s|}{2}$ and for this $\epsilon$ we have a natural number $n_0$ such that $n≥n_0$ implies $|s_n-s|<\epsilon$ i.e. in particular for all $n≥max(n_0,k+1)$ we have $s_n-s≤|s_n-s|<\epsilon$ i.e. $s_n<s+\frac{|s|}{2} =s-\frac{s}{2}<0$ for all $n≥max(n_0,k+1)$, a contradiction since $s_n>0$ for $n>k$.

Answer 3

Ok so let me see if I can combine this with what I was thinking: Suppose that the limit, s, s.t. $s<0$ then it follows by the definition of a limit that $\exists$ some N s.t. $|s-s_n|<1/n$ and for some $n>N$ and given that $s_n>0$ and $s<0$ then $|s-s_n|=s_n-s<1/n$ further $-s<s_n-s<1/n$ (again since $s<0$); this imples $-s<1/n$, but since 1/n is arbitrarily small, this is a contradiction and so $s\ge0$

Is this right?