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.

[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$.
