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Can you say that $\limsup a_n$ is the upper bound of the sequence $a_n$ and $\liminf a_n$ is the lower bound of $a_n$?

Sebastiano
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TheAsker
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  • Consider the sequence $5+1, 3-1, 5+1/2, 3-1/2, 5+1/3, 3-1/3...$. The limsup is $5$ and yet the upper bound is $6$. Similarly the liminf is $3$, and yet the lower bound is $2$ – Vercingetorix Dec 08 '20 at 06:54
  • Not necessarily. Consider the sequence $x_n=\frac{1}{n}$. Note that the $\limsup x_n=\liminf x_n=\lim x_n=0$, but for $x_1>\limsup x_n$. – Ryan Dec 08 '20 at 06:55

1 Answers1

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No.

If $a_n=(-1)^n + \frac{(-1)^n}{n}$, then

  1. the upper bound of the sequence is $\frac{3}{2}$
  2. the lower bound for the sequence is $-2$
  3. $\limsup a_n = 1$
  4. $\liminf a_n = -1$

so you have a case when the $\limsup$ is not the upper bound, and the $\liminf$ is not the lower bound.

5xum
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  • mm and what If An->INFINITY or An->-INFINITY – TheAsker Dec 08 '20 at 06:58
  • @TheAsker I suggest you give it a little thought first. What do you think happens then? – 5xum Dec 08 '20 at 06:59
  • @TheAsker Also, please read the following before writing mathematical equations: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 5xum Dec 08 '20 at 06:59
  • If An->INFINITY than limsupAn->INFINITY but you already proved I was wrong once >< – TheAsker Dec 08 '20 at 07:00
  • @TheAsker $\limsup a_n \to\infty$ is not the correct way to write it. It's $\limsup a_n = \infty$. But yes, you are correct, if $a_n$ diverges to $\infty$, then $\limsup a_n=\infty$. But I don't see what that has to do with the original question. Also, please use mathjax. – 5xum Dec 08 '20 at 07:02