# Prove that limit inferior is same as limit superior for a convergent sequence

I was reading the book "Understanding Analysis" by Stephen Abbott on my own. I came across the following problem. Let $(a_n)$ be a convergent sequence.
Let $y_n$=sup{$a_k:k\geq n$}. Then lim sup $a_n$ = lim $y_n$.
Similarly define lim inf $a_n$.
Prove that lim inf $a_n$ = lim sup $a_n$ if and only if $(a_n$) converges and in that case all three share the same value.
I don't know how to prove it. I know that there are alternate definitions of lim sup and lim inf . So if anyone uses those definitions in their proof, please mention those clearly and if possible, also state how that definition is equivalent to the above definition . A proof which doesn't use metric spaces etc is preferred since I have not studied it yet.

• If $\lim\sup$ and $\lim\inf$ are distinct, then there are subsequences of $(a_n)$ which converge to distinct points. This means that $(a_n)$ isn't a Cauchy sequence and therefore isn't a convergent sequence. Commented Dec 26, 2014 at 11:32
• Ok. It's that simple! Thankyou. Well this question is given in the book even before it is explained what is a limit point. So I assume there would be a solution just using the basic definition of convergence Commented Dec 26, 2014 at 11:37
• Also, the proof for the converse can be deduced from the fact that $\lim\inf a_n\leq \lim a_n\leq\lim\sup a_n$, which can be obtained from the basic definition of $\lim\inf$ and $\lim\sup$. Commented Dec 26, 2014 at 11:42

[As in the problem in Abbott's book, we will assume only that $(a_n)$ is a bounded sequence.]

For all $N\in\mathbb{N}$, define $v_{N}=\sup\{s_n:n\ge N\}$ and $u_{N}=\inf\{s_n:n\ge N\}$; so by definition

$\limsup a_n=\displaystyle\lim_{N\to\infty}v_N$ and $\liminf a_n=\displaystyle\lim_{N\to\infty}u_N$.

$\textbf{1)}$ Suppose $\limsup a_n=\liminf a_n=L$, and let $\epsilon>0$ be given.

a) Since $\displaystyle\limsup a_n=\lim_{N\to\infty}v_N=L, \;\;v_N<L+\epsilon$ for some $N\in\mathbb{N} \;\text{ so }a_n<L+\epsilon \text{ for }n\ge N$.

b) Since $\displaystyle\liminf a_n=\lim_{N\to\infty}u_N=L, \;\;L-\epsilon<u_M$ for some $M\in\mathbb{N} \;\text{ so }L-\epsilon<a_n\text{ for }n\ge M$.

If $K=\max\{N,M\}$, then $L-\epsilon<a_n<L+\epsilon \text{ for }n\ge K;$ $\;\;$so $\displaystyle\lim_{n\to\infty}a_n=L$.

$\textbf{2)}$ Suppose $\displaystyle\lim_{n\to\infty}a_n=L$, and let $\epsilon>0$ be given.

Then there is an $N\in\mathbb{N}$ such that $|a_n-L|<\epsilon$ for $n\ge N$, so $L-\epsilon<a_n<L+\epsilon$ for $n\ge N$.

Therefore $L-\epsilon\le u_N \text { and }v_N\le L+\epsilon$, so $L-\epsilon\le \liminf a_n \text{ and }\limsup a_n\le L+\epsilon$

$\hspace{2.7 in}$since $(u_N)$ is increasing and $(v_N)$ is decreasing.

Since $\epsilon>0$ was arbitrary, $\;\;$$L\le\liminf a_n\le \limsup a_n\le L \; so \;$$\liminf a_n=\limsup a_n=L$.

Notice that $\limsup_n a_n = \ell$ is equivalent to say that:

• for all $\varepsilon>0$ eventually one has $a_k < \ell + \varepsilon$
• for all $\varepsilon>0$ frequently one has $a_k > \ell -\varepsilon$

where eventually means that the following property is true for all sufficiently large $k$ while frequently means that the following property is true for arbitrary large values of $k$ (I'm not sure the english translation is the correct one).

Formally: we say that $P(k)$ holds eventually if there exists $n$ such that $P(k)$ holds for all $k\ge n$. We say that $P(k)$ holds frequently if for all $n$ there exists $k\ge n$ such that $P(k)$ holds.

The definition of $\lim_n a_n = \ell$ is the same as above with frequently in the second line replaced by eventually while the definition of $\liminf_n a_n=\ell$ has frequently and eventually exchanged.

Notice that eventually implies frequently and you are done...

• Nice answer - this made this alternate definition a little clearer for me. (Abbott uses "eventually" instead of "definitely" in an exercise in his book.) Commented Dec 26, 2014 at 19:53
• I've corrected it. I was sure my translation from italian was not good... Commented Dec 27, 2014 at 7:00

First we prove $\displaystyle \lim_{n \to \infty} y_n$ exists. Assume that $\{a_n\}$ is a convergent sequence to begin with, then it is bounded, and this implies that $\{y_n\}$ is also bounded below in particular. And $\{y_n\}$ is a decreasing sequence, thus Bolzano's lemma says it converges to a finite limit. Call it $y$, and call the limit of the sequence $\{a_n\}$ $x$. We show $x = y$. Let $\epsilon > 0$ be arbitrary, there exists indices $k_0,n_0, m_0 \in \mathbb{N}$ such that: $k_0 \geq m_0+n_0$, and:

$|y-y_n| < \dfrac{\epsilon}{3}, n \geq n_0$,

$|a_n-x| < \dfrac{\epsilon}{3}, n \geq m_0$,

$|y_{n_0+m_0} - a_{k_0}| = y_{n_0+m_0} - a_{k_0} < \dfrac{\epsilon}{3}$.

Thus:
$|y-x| \leq |y-y_{m_0+n_0}| + |y_{m_0+n_0} - a_{k_0}| + |a_{k_0} - x| < \dfrac{\epsilon}{3} + \dfrac{\epsilon}{3} + \dfrac{\epsilon}{3} = \epsilon$, and since this is true for all $\epsilon > 0$, we deduce that : $|y-x| = 0 \Rightarrow x = y$.

The rest of the work is done in a similar fashion.

There is a concept called adherence value. A real number is an adherence value of a sequence if any neighborhood of this real number contains infinitely many terms of the sequence. Note that you can define this also for $\pm\infty$ in which case it simply means that the sequence has infinitely many arbitrary big positive (or negative) values.

Now, it's easy to see that lim sup (resp lim inf) of a sequence is the biggest (resp. the smallest) adherence value of the sequence. Now, saying that they're equal is equivalent to say that the sequence has only one adherence value. It is well known that this implies that in fact the sequence converges to this unique adherence value.