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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.

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    $\begingroup$ Assuming that $\lim a_{n} =s$, then, for every $\epsilon > 0$, there exists $N \ge 1$ such that $s-\epsilon/2 < a_{n} < s+\epsilon/2$ whenever $n \ge N$. Therefore $s-\epsilon < b_{n} < s+\epsilon$ for $n \ge N$, where $b_{n}=\sup\{ a_{k} : k \ge n \}$. $\endgroup$ – DisintegratingByParts Feb 7 '14 at 5:53
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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 $\lim\, a_{n} = s$.

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  • $\begingroup$ The $\epsilon/2$ (probably copied from @T.A.E.'s comment) is unnecessary: if $s-\epsilon\leqslant s_n\leqslant s+\epsilon$ for every $n\geqslant N$ then $s-\epsilon\leqslant a_n\leqslant s+\epsilon$ for every $n\geqslant N$. $\endgroup$ – Did Feb 8 '14 at 8:16
  • $\begingroup$ @Did: I had first written the same as u mention in comment, but later edited to add $\epsilon/2$ just to conform to the usual limit definition which uses $<$ instead of $\leq$ (although both forms are equivalent). $\endgroup$ – Paramanand Singh Feb 8 '14 at 15:08

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