$f \in O(g)$, iff $\limsup_{n\to\infty}|\frac{f(n)}{g(n)}| < \infty $ Given the functions $f$, $g$: $\mathbb{N} \to \mathbb{R}$
I have to prove that,
$f \in O(g)$, iff $\limsup_{n\to\infty}|\frac{f(n)}{g(n)}| &lt \infty  $
How can I prove it formally (at best using the O-Notation definitions)?
Sorry to pose the question quite straightforward without describing what
I already tried and where I got stuck. In this case, I just have no clue where to start from.
Maybe a hint of anything to start with should suffice.
Thanks!
 A: Let's write the definition of $f \in O(g(n))$ and $\lim\sup_{n\rightarrow\infty}|\frac{f(n)}{g(n)}|&lt\infty$. We get
\begin{align}
(1)& \lim_{n\rightarrow\infty}\left(\sup\left\{\left|\frac{f(k)}{g(k)}\right| : k \ge n\right\}\right) &lt \infty \\
(2)& \ f(n)\in O(g(n)) :\Longleftrightarrow \text{$\exists N,K>0$ $\forall n\ge N\colon\ |f(n)| \le K\ |g(n)|$.}
\end{align}
We define Let $S_n := \sup\left\{ \left|\frac{f(k)}{gk)}\right|: k\ge n\right\}$.
(1)$\Rightarrow$(2): Let $a$ be the limit to which the sequence in (1) converges. By definition of the limit we have that
$\forall \varepsilon > 0\ \exists N\ \forall n\ge N\colon | S_n - a | \le \varepsilon.$ Fix any $\varepsilon>0$ and set $K=\varepsilon+a$. The above implies that there exists an $N>0$ such that $|f(n)| &lt (\varepsilon+a)|g(n)|$ for all $n\ge N$, which matches the right hand side of (2). 
(2)$\Rightarrow$(1):  Starting from the right hand side of (2), we immediately see that $S_N \le K$ and in fact $S_n \le K$, for any $n \ge N$. We observe that $(S_n)_{(n\ge N)}$ is an increasing sequence that is bounded from above by $K$. Thus $\lim_{n\rightarrow\infty}S_n$ is converging, which shows (1). 
