Proving that the $\lim_{n\to \infty} a_n = \infty$ where $\lim_{n \to \infty} \frac{a_n + 1}{a_n} = k > 1$ I am working on a proof as follows: Let $k > 1$ and $(a_n)_{n \in \Bbb{N}}$ be a sequence where $a_n > 0$ for all $n \in \Bbb{N}$. $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = k$$ Show that $\lim_{n \to \infty} a_n = +\infty$ I have ascertained that when n is large enough, $\frac{a_{n+1}}{a_n} > 1$ and that $a_{n+1} > a_n$ so $(a_n)$ will be strictly monotone increasing. However, this is not strong enough to show the sequence tends to infinity. I feel like I understand the notion of why it tends to infinity, the strict inequality meaning it won't converge however I'm not sure how to prove it. Any help will be greatly appreciated.

> Herein, we present a standard approach that relies on the definition of the limit. To that end we proceed.

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If $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=k>1$, then for all $\epsilon>0$, there exists a number $N(\epsilon)$ such that whenever $n>N(\epsilon)$,

$$k-\epsilon<\frac{a_{n+1}}{a_n}
Take $\epsilon=\frac{k-1}{2}$. Then, for $n>N\left(\frac{k-1}{2}\right)$, we see that

$$a_{n+1}>\left(\frac12+\frac k2\right)a_n$$

where $\left(\frac12+\frac k2\right)>1$.

Proceeding recursively, we find that

$$a_{n+m}>\left(\frac12+\frac k2\right)^ma_n$$

Letting $m\to \infty$, we obtain the coveted limit

$$\lim_{n\to \infty}a_n=\infty$$

And we are done!