Suppose that $\lim_{n→∞} \frac{S_n}{n} > 0$ Show that $S_n → ∞$. ($S_n$ is a sequence) So at first glance it appeared pretty easy, we can assume by contradiction that sn converges to some constant L, then $\lim(\frac{S_n}{n})$ as n approaches infinity is $\lim(\frac{L}{n})$ which is zero hence a contradiction (0>0 wrong statement). but after thinking about it $\lim(\frac{L}{n})$ approaches zero but never exactly zero,so the statement $\lim(\frac{L}{n})>0$ is not wrong. am i overthinking this?
 A: Yes we could prove by contradiction considering all the cases:


*

*$S_n\to l \implies \frac{S_n}n \to 0$

*$S_n\to -\infty \implies \frac{S_n}n \to m$ with $m\le 0$ (to prove better with detail)

*limit doesn't exists (we need to consider $\limsup$ and $\liminf$)
As a simpler alternative, by definition we have that
$$S_n/n \to L>0 \implies \forall \epsilon \quad\exists n_0\quad \forall n>n_0\quad |S_n/n-L|<\epsilon$$
that is, assuming $\epsilon=L/2$, $\forall n>n_0$ we have 
$$S_n/n> L/2 \implies S_n>n\cdot L/2$$
A: Let $0<L=\lim_{n\to \infty}S_n/n.$ For any $e>0$ there exists $n_e\in \Bbb N$ such that $n\ge n_e\implies |S_n/n-L|<e.$ In particular, with $e=L/2$ we have $$n\ge n_{L/2}\implies |S_n/n-L|<L/2\implies$$ $$\implies S_n/n-L\ge -L/2\implies$$ $$\implies S_n/n>L/2\implies$$ $$\implies S_n>n(L/2).$$
Now, given any $x\in \Bbb R,$ take $m_x\in \Bbb N$ such that $m_x(L/2)>x.$ And let $p_x=\max (n_{L/2},m_x).$ Then  $$n\ge p_x \implies S_n>n(L/2) \quad \text {(because $n\ge n_{L/2}$)}$$ $$ \ge m_x(L/2)\quad \text {(because $n\ge m_x$)}$$ $$>x.$$
A: The statement in your final sentence is wrong. lim(L/n) = 0, it's not >0. Just because the members of the sequence are "never exactly zero", this doesn't mean the limit cannot equal zero (which is does).
