Limit, Limsup clarification needed.

Question

I know that if $(a_n)$ and $(b_n)$ are convergent sequences, and $a_n \leq b_n$, $\forall n\ge N$, then $\lim\, a_n\leq \lim\,b_n$. Now my question is what if we don't know that $(a_n)$ and $(b_n)$ are convergent? Specifically, my question relates to a problem in my analysis text in which we assume a sequence $(s_n)$ converges to $s$ and try to show that $\limsup\,s_n$ converges to $s$. It defines $a_n = \sup_{k\ge n}(a_k: a_k\in (a_k))$, and shows $a_n \leq s + \epsilon, \forall n\ge N$, where $s$ and $\epsilon > 0$ are fixed real numbers. Then it claims $\lim\,a_n\le \lim\,(s+\epsilon)=s+\epsilon$. I know that $\lim\,a_n$ always exists in the extended $\mathbb{R}$, but is there some way to see that it actually converges here. Any clarification would be appreciated.

Answer 1

I assume that the question is as follows:

Let $\{s_{n}\}$ be a sequence converging to $s$ and let $a_{n} = \sup_{k \geq n}\{s_{k}\}$. Show that $\lim\, a_{n} = s$.

Note that there is a slight mistake in the question from OP where he tries to define $a_{n} = \sup_{k \geq n}\{a_{k}: a_{k} \in (a_{n})\}$. You can't have the same symbol $a_{n}$ both sides here.

We will show that $\{a_{n}\}$ converges to $s$. Let $\epsilon > 0$ be arbitrary. Since $s_{n} \to s$ as $n \to \infty$, it follows that there is a positive integer $N$ such that $s - (\epsilon/2) < s_{n} < s + (\epsilon/2)$ for all $n \geq N$. It now follows from the definition of $a_{n}$ that $$s - \epsilon < s - (\epsilon/2) \leq a_{n} \leq s + (\epsilon/2) < s + \epsilon$$ for all $n \geq N$. Hence $\lim\, a_{n} = s$.