Trying to understand the definition of limit superior Given sequence ($a_{n}$), I know the definition of limit superior of ($a_{n}$). I also understood the geometric meaning of term which actually says that it is the largest of all accumulation points of sequence. But I never understood how this definition is actually going to give me this thing. 
I am adding definition : 
 Given ($a_{n}$) 
put $b_{k}$ = sup {$a_{k}, a_{k+1},$ ...} (k=1,2,3,...)
and $\beta$= inf{$b_{1}, b_{2},$...}
Then $\beta$= lim sup of ($a_{n}$). Thanks.
 A: Denote $l=\limsup_{n}a_{n}$, where $\limsup$ is defined using your
definition. Then, we have three cases. Case 1: $l\in\mathbb{R}$,
Case 2: $l=-\infty$, Case 3: $l=\infty$.
Firstly, let us consider Case 1. We assert that $l$ is the greatest accumulation
point of $(a_{n})$ by proving the following facts:


*

*For each $\varepsilon>0$, there are at most finitely many $n$
such that $a_{n}\in(l+\varepsilon,\infty)$.

*For each $\varepsilon>0$, there are infinitely many $n$ such
that $a_{n}\in(l-\varepsilon,l+\varepsilon)$.
Condition 1 tells us that the sequence $(a_{n})$ has no accumulation
point in the region $(l,\infty)$. Condition 2 tells us that $l$
is an accumulation point of the sequence $(a_{n})$. Combining, we
conclude that $l$ is the largest accumulation of the sequence $(a_{n})$.
We go to prove (1): Prove by contradiction. Suppose that there exists
$\varepsilon_{0}>0$ such that there are infinitely many $n$ satisfying
$a_{n}\in(l+\varepsilon_{0},\infty)$. For each $k\in\mathbb{N}$,
$b_{k}$ (defined by you) satisfies $b_{k}\geq l+\frac{1}{2}\varepsilon_{0}$.
Therefore, $l=\lim_{k}b_{k}\geq l+\frac{1}{2}\varepsilon_{0}$, which
is a contradiction.
We go to prove (2): Let $\varepsilon>0$ be given. Since $l=\lim_{k}b_{k}$,
there exists $K_{1}\in\mathbb{N}$ such that $b_{k}\in(l-\varepsilon/2,l+\varepsilon/2)$
whenever $k\geq K_{1}$. By condition $(1)$, we can choose $K_{2}\in\mathbb{N}$
such that $a_{n}\notin(l+\frac{1}{2}\varepsilon,\infty)$ whenever
$n\geq K_{2}$. Let $K=\max(K_{1},K_{2})$. Since $b_{K}>l-\frac{1}{2}\varepsilon$,
$l-\frac{1}{2}\varepsilon$ is not an upper bound of the set $\{a_{K},a_{K+1},\ldots\}$.
There exists $n_{1}\geq K$ such that $a_{n_{1}}>l-\frac{1}{2}\varepsilon$.
Since $b_{n_{1}+1}>l-\frac{1}{2}\varepsilon$, there exists $n_{2}\geq n_{1}+1$
such that $a_{n_{2}}>l-\frac{1}{2}\varepsilon$. Note that $n_{1}<n_{2}$.
Continuing the process indefinitely (formally, by applying the Recursion
Theorem, which is often mixed-up with induction...), we obtain a subsequence
$(a_{n_{k}})_{k}$ such that $l-\frac{1}{2}\varepsilon<a_{n_{k}}$
for all $k=1,2,\ldots$. Since $n_{k}\geq K_{2}$ for all $k$, we
have $a_{n_{k}}\leq l+\frac{1}{2}\varepsilon$. In short, $a_{n_{k}}\in(l-\varepsilon,l+\varepsilon)$
for all $k$.
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Case 2: In this case, we can prove that $\lim_{n}a_{n}=-\infty$ (left
as exercise)
Case 3: In this case, we can prove that there exists a subsequence
$(a_{n_{k}})$ such that $a_{n_{k}}\rightarrow\infty$ (left as exercise)
