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

Solution 1:

[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$.

Solution 2:

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...