Is this proof about limit of a sequence correct? I have to prove that if $f_n$ is convergent and $g_n$ is divergent to positive infinity then show that $lim\frac{f_n}{g_n}$=$0$.
My attempt: Since $f_n$ is convergent, it is bounded.
This implies, $|f_n|\leq G\epsilon$.
And since $g_n$ is divergent, this implies that corresponding to a pre-assigned positive number G, however large, there exists a natural number K such that
$g_n>G$ $ \forall n\geq K$
Consider, $|{\frac{f_n}{g_n}-0}|\leq \frac{|f_n|}{|g_n|} <\frac{G\epsilon}{G}<\epsilon$
Thus,  $lim\frac{f_n}{g_n}$=$0$.
Is it right or wrong? Thanks in advance.
 A: The statement might be not true if $f_n$ converges to $0$. Consider $f_n =1/n$ and $g_n = n$.
Otherwise, by assuming $f_n \not \to 0$, there are several flaws in your proof.
When you prove sequence convergence using $\varepsilon$ definition, you start from some $\varepsilon$ and you prove that for a large enough rank $N$, all the term of higher rank are closer to the converging value than $\varepsilon$. $(f_n)$ is convergent and so it is bounded, as you noted. However since $f_n \not \to 0$ you cannot write $|f_n| \le G \varepsilon$ unless $G$ and $\varepsilon$ move around at the same time, not proving anything in the end as $G$ is not fixed at all as you assumed it was. You can only say that $|f_n| \le M$ for some constant $M$. Actually, $G$ is the important parameter of your proof, playing the role of $1/\varepsilon$.
So here is a proof proposition :
$g_n \to \infty$ implies $1/g_n \to 0$, thus :
$$\forall \varepsilon >0, \exists N, \forall n>N, |1/g_n| <\varepsilon$$
Assume $|f_n| \le M$ and let $\varepsilon >0$.
Thus, for $n>N$:
$$\left| \frac {f_n}{g_n} \right|\le \left| {f_n} \right|\varepsilon \le M \varepsilon.$$
Now let $\varepsilon' =  M \varepsilon$. Since $\varepsilon$ is arbitrary, $\varepsilon'$ is too. So for any $\varepsilon'>0$ and for $n$ large enough, $\left| \frac {f_n}{g_n}\right| \le \varepsilon'$ and the sequence $f_n/g_n$ indeed converges to $0$.
