Suppose $(a_n)_{n\ge1}$ is a positive (so non-zero), real sequence with $\lim_{n\rightarrow\infty}a_n=L$, where $L\in[0,\infty]$. Is this equivalent with $\limsup_{n\rightarrow\infty}a_n=\liminf_{n\rightarrow\infty}a_n=L$? In particular, does this hold for the infinity case? If so, why? Thank you!
[Math] When does a limit of a sequence equal both its lim sup and lim inf
limitslimsup-and-liminfreal-analysissequences-and-series
Related Solutions
The $\limsup$ is the largest real number, or $\pm\infty$, which is a limit of a subsequence of $S_n$, whereas $\liminf$ is the smallest real number (or $\pm\infty$) which is a limit of a subsequence of $S_n$.
It follows that $\lim a_n$ exists (in the broad sense) if and only if $\liminf a_n=\limsup a_n$.
The existence of $\liminf,\limsup$ follows from the completeness of $\mathbb R$.
If $S_n$ is a sequence such that $\limsup S_n=\infty$ then there is a subsequence, $S_{n_k}$ which is strictly increasing. Therefore its limit is $\infty$.
To the edit:
One cannot treat $\infty$ as a real number. It's good when a sequence has a limit within the real numbers, but sometimes it is sufficient that it is convergent, i.e. satisfies some definition.
To say that the limit of a sequence is $\infty$ is to say that although the sequence does not tend to a real number, it behaves "nice enough". This is in contrast to sequences which have no limit at all and just jump around between numbers.
This is why we sometimes say that it converges in a broad sense when it has the limit $\pm\infty$. Sometimes, when context is clear enough, we may omit the "broad sense" part too.
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 $x$.
Best Answer
The $\displaystyle\liminf_{n \to +\infty} a_n$ denotes the smallest adherence value of the sequence $(a_n)_n$. The $\displaystyle\limsup_{n \to +\infty} a_n$ denotes the greatest adherence value of the sequence $(a_n)_n$. Recall that $\ell$ is an adherence value of $(a_n)_n$ if there is a subsequence of $(a_n)_n$ which converges to $\ell$. Thus, if the limit inf is equal to the limit sup then the whole series converges to this common limit. The reverse is also true.