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Suppose $x_n$ is a bounded sequence and $\limsup x_n = \liminf x_n = c$. Prove $x_n \rightarrow c, n \rightarrow \infty$

Not sure if there is a better way to show this, or if my way is even correct, just looking for some guidance or tips.

Since $\limsup x_n = c$, then given $\epsilon > 0$, there exists a $N_1$ such that for all $n \geq N_1$ we have:

$x_n < c + \epsilon$.

Likewise, since $\liminf x_n = c$, then there exists some $N_2$ such that for all $n \geq N_2$ we have:

$c - \epsilon < x_n$

Let $N = \max\{N_1, N_2\}$, then for all $n \geq N$, we get:

$c - \epsilon < x_n < c + \epsilon$

$|x_n - c| < \epsilon$

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Your proof is correct. You might write down the definition of $\limsup$ before concluding that $x_n<c+\epsilon$ and $x_n>c-\epsilon$.

$$\limsup_{n\to\infty}x_n=c\iff\forall \epsilon>0\ \exists N_1\in\mathbb N\ \forall n\ge N_1, -\epsilon<\sup\{x_n,x_{n+1}\ldots\}-c<\epsilon$$

After this, you may safely conclude that $x_n<c+\epsilon$ for $n\ge N_1$

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  • $\begingroup$ I'm sorry, not quite sure I understand what I should do? $\endgroup$ – student_t Jan 29 '17 at 21:51
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    $\begingroup$ I explained in more detail. $\endgroup$ – Momo Jan 29 '17 at 21:57
  • $\begingroup$ Oh I see, thanks! Quick question, isn't: $-\epsilon < \sup\{x_n, x_{n+1}, ...\} -c$ true for all n? $\endgroup$ – student_t Jan 29 '17 at 22:03
  • $\begingroup$ Right, because $\sup\{x_n,x_{n+1}\ldots\}$ decreases to $c$. $\endgroup$ – Momo Jan 29 '17 at 22:09

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