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Show that if the sequence$(x_n)$ is bounded, then $(x_n)$ converges iff $\lim\sup_{x\to\infty}(x_n)=\lim\inf_{x\to\infty}(x_n)$.

The definitions that I’m using:

$$\begin{align*} &\liminf_{n\to\infty}x_n=\lim_{n\to\infty}\inf_{m\ge n}x_m\\ &\limsup_{n\to\infty}x_n=\lim_{n\to\infty}\sup_{m\ge n}x_m \end{align*}$$

This’s the first time I deal with $\lim\inf$ things, can someone give me help?

Thank you.

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There are at least two possible definitions of $\limsup$ and $\liminf$ that you might have been given; what definitions are you using? –  Brian M. Scott Jun 29 '13 at 7:20
    
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Here's a little help: give names to the sequences $\inf_{m\geq n} x_m$ and $\sup_{m\geq n} x_m$. How can you compare those to $x_n$ ? What does it mean that they have same limit ? –  zozoens Jun 29 '13 at 7:46
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1 Answer

up vote 2 down vote accepted

I’ll get you started. For one direction, suppose that $\lim\limits_{n\to\infty}x_n=x$; we want to show that $$\limsup_{n\to\infty}x_n=\liminf_{n\to\infty}x_n\;.$$ The most natural guess is that this is true because both are equal to $x$, so let’s try to prove that.

In order to show that $\limsup\limits_{n\to\infty}x_n=x$, we must show that $\lim\limits_{n\to\infty}\sup_{k\ge n}x_k=x$. To do this, we must show that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that

$$\left|x-\sup_{k\ge n}x_k\right|<\epsilon\quad\text{whenever}\quad n\ge m_\epsilon\;.$$

Since $\lim\limits_{n\to\infty}x_n=x$, what we actually know is that for each $\epsilon>0$ there is an $m_\epsilon'\in\Bbb N$ such that $|x-x_n|<\epsilon$ whenever $n\ge m_\epsilon'$.

  • Show that if $|x-x_n|<\epsilon$ for all $n\ge m_\epsilon'$, then $\left|x-\sup\limits_{k\ge n}x_k\right|\le\epsilon$. Conclude that if we set $m_\epsilon=m_{\epsilon/2}'$, say, then $$\left|x-\sup_{k\ge n}x_k\right|<\epsilon\quad\text{whenever}\quad n\ge m_\epsilon$$ and hence $\limsup\limits_{n\to\infty}x_n=x$.

  • Modify the argument to show that $\liminf\limits_{n\to\infty}x_n=x$.

For the other direction, suppose that $$\limsup_{n\to\infty}x_n=\liminf_{n\to\infty}x_n=x\;;$$ we want to show that $\langle x_n:n\in\Bbb N\rangle$ converges. The natural candidate for the limit of the sequence is $x$, so we should try to prove that $\lim\limits_{n\to\infty}x_n=x$, i.e., that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|x-x_n|<\epsilon$ whenever $n\ge m_\epsilon$. What we know is that

$$\lim_{n\to\infty}\sup_{k\ge n}x_k=x=\lim_{n\to\infty}\inf_{k\ge n}x_k\;,$$

i.e., that for each $\epsilon>0$ there is an $m_\epsilon'\in\Bbb N$ such that

$$\left|x-\sup_{k\ge n}x_k\right|<\epsilon\quad\text{and}\quad\left|x-\inf_{k\ge n}x_k\right|<\epsilon\quad\text{whenever}\quad n\ge m_\epsilon'\;.$$

(Why can I use a single $m_\epsilon'$ instead of requiring separate ones for each of the two limits?)

  • Show that if $\ell\ge n$, then $$|x-x_\ell|\le\max\left\{\left|x-\sup_{k\ge n}x_k\right|,\left|x-\inf_{k\ge n}x_k\right|\right\}\;,$$ and conclude that setting $m_\epsilon=m_\epsilon'$ will ensure that $|x-x_n|<\epsilon$ whenever $n\ge m_\epsilon$ and hence that the sequence converges to $x$.
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"Show that if $|x-x_n|<\epsilon$ for all $n\ge m_\epsilon'$, then $\left|x-\sup\limits_{k\ge n}x_k\right|\le\epsilon$." This seems trivial, do I really need to write something about it? –  ᴊ ᴀ s ᴏ ɴ Jun 30 '13 at 6:19
    
@ᴊᴀsᴏɴ: I grant that it’s intuitively obvious, but it’s not entirely trivial to write down a justification, so yes, you ought to do so. –  Brian M. Scott Jun 30 '13 at 6:22
    
So actually for this part, we first say $|x_n-x|\le\epsilon/2,\forall n\geq N$, then conclude that $|\sup_{k\geq n}x_k-x|\le\epsilon,\forall n\geq N$, right? –  ᴊ ᴀ s ᴏ ɴ Jun 30 '13 at 6:31
    
@ᴊᴀsᴏɴ: For the first part, yes. –  Brian M. Scott Jun 30 '13 at 6:37
    
In the second part, can the "$\ell\geq n$" be changed to "$\ell\geq m_\epsilon'$"? –  ᴊ ᴀ s ᴏ ɴ Jun 30 '13 at 6:46
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