1. ## Sequence question

I am studying for my final in undergrad analysis, and have a question concerning sequences. Here it is. Let $\displaystyle \{s_n\}$ be a sequence $\displaystyle \backepsilon \{s_n\}>0 \; \forall \; n \in \mathbb{N}.$ Suppose $\displaystyle \lim_{n \to \infty}s_n=s.$ Show that $\displaystyle s \ge 0.$

2. Pf: Assume $\displaystyle s_n > 0 \forall n \in \mathbb{N}$ and $\displaystyle \displaystyle\lim_{n\to\infty} s_n = s$ Now assume $\displaystyle s < 0$ So, By def $\displaystyle \forall \epsilon > 0 \exists N \in \mathbb{N} | \forall n \geq N \: |s_n - s|<\epsilon$ By the triangle ineq $\displaystyle \forall \epsilon > 0 \: |s_n| - |s| < \epsilon$ which means $\displaystyle |s_n| \rightarrow |s| as n \rightarrow \infty$ since $\displaystyle s_n > 0 \forall n$ we have $\displaystyle s_n \rightarrow |s| as n \rightarrow \infty$ since $\displaystyle |s|>0$ this is a contradiction.
3. In other words, if s<0, then for a given $\displaystyle \epsilon$ with $\displaystyle s+\epsilon<0$, you should have elements of $\displaystyle s_n$ existing in the $\displaystyle \epsilon$ neighborhood of s (by limit to s). Since $\displaystyle s+\epsilon<0$, then those elements are also negative which gives a contradiction.
4. I think there's a somewhat easier way. Suppose that $\displaystyle s_{n} > 0$ for all $\displaystyle n\in \mathbb{N}$. Suppose $\displaystyle \lim_{n\to \infty} s_{n} = s < 0$. Then we know there exists $\displaystyle N$ such that $\displaystyle n>N$ implies $\displaystyle |s_{n} - s| < -s$. This is true because such an implication must hold for any $\displaystyle \epsilon > 0$ and if $\displaystyle s < 0$ then $\displaystyle -s > 0$, so this is the particular case where $\displaystyle \epsilon = -s$. So this means $\displaystyle s < s_{n} - s < -s$ for $\displaystyle n>N$. Adding $\displaystyle s$ to each part of the equality, we get for $\displaystyle n>N$, $\displaystyle 2s < s_{n} < 0$. That means after some point, $\displaystyle s_{n}$ will be negative! And there arises your contradiction.