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.
 A: Choose an $\epsilon > 0$ such that $1+\epsilon < k$. We know that there exists an $N$ such that
$$\frac{a_{n+1}}{a_n}>1+\epsilon$$
for all $n\ge N$. Thus we have that $a_{n+1}>(1+\epsilon)a_n$. Then applying this recursively we have that
$$a_{N+k}>(1+\epsilon)^ka_N$$
And we see that $a_N>0$ and $(1+\epsilon)^k\to\infty$ so $a_{N+k}\to\infty$.
A: 
Herein, we present a standard approach that relies on the definition of the limit.  To that end we proceed.


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 $n>N(\epsilon)$, 
$$k-\epsilon<\frac{a_{n+1}}{a_n}<k+\epsilon$$
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!
A: Hint
there exists $a>1$ and $N\geq 0$  s.t
for $n\geq N$,
$$\frac{u_n}{u_{n-1}}\geq a$$
$$\frac{u_{n-1}}{u_{n-2}}\geq a$$
.
.
$$\frac{u_{N+1}}{u_N}\geq a$$
use telescoping and geometric sequence.
A: lim{a_(n+1)}=lim {a_(n+1)/a_n}*{a_n}=lim{a_(n+1)/a_n}*lim{a_n}. This only works if lim a_n is infinity. 
