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:

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

$|x_n - c| < \epsilon$


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 $n\ge N_1$

  • $\begingroup$ I'm sorry, not quite sure I understand what I should do? $\endgroup$ – student_t Jan 29 '17 at 21:51
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.