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I wish to prove the statement.

If $\lim_{n\to\infty}{\frac{s_n}{n}} >0$, then $s_n\to\infty$.

I will begin by assuming that $\lim_{n\to\infty}{\frac{s_n}{n}}=L$. Then I know that given $\epsilon>0$, $\exists N\in \mathbb{N}$ such that $|\frac{s_n}{n}-L|<\epsilon$ for all $n\geq N$.

Im not sure where to go from here. Any hints?

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Let $L:=\lim_{n\rightarrow\infty}\dfrac{s_{n}}{n}>0$, then for some $N$, we have $\left|\dfrac{s_{n}}{n}-L\right|<\dfrac{L}{2}$ for all $n\geq N$, so $\dfrac{s_{n}}{n}>L-\dfrac{L}{2}=\dfrac{L}{2}$ for all such $n$, then given any $M>0$, for all $n\geq\dfrac{2M}{L}+N$, then $s_{n}=\dfrac{s_{n}}{n}\cdot n>\dfrac{L}{2}\cdot n>\dfrac{L}{2}\cdot\dfrac{2M}{L}=M$, this shows that $\lim_{n\rightarrow\infty}s_{n}=\infty$.

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  • $\begingroup$ Actually we want those $n$ greater than both $\dfrac{2M}{L}$ and $N$, for the former, then we are allowed to use $n>\dfrac{2M}{L}$, so that $\dfrac{L}{2}\cdot\dfrac{2M}{L}=M$ is perfect ending, for the latter, we are allowed to use $\dfrac{s_{n}}{n}>\dfrac{L}{2}$. $\endgroup$ – user284331 Feb 24 '18 at 3:40

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