Show that this limit is equal to $\liminf a_{n}^{1/n}$ for positive terms.

Show that if $a_{n}$ is a sequence of positive terms such that $\lim\limits_{n\to\infty} (a_{n+1}/a_n) $ exists, then this limit is equal to $\liminf\limits_{n\to\infty} a_n^{1/n}$.

I am not event sure where to start from, any help would be much appreciated.

The typical textbook proof goes as follows.

If the $\lim \frac{a_{n+1}}{a_n} = q \gt 0$, then given an $\epsilon \gt 0$ such that $q \gt \epsilon$, there is some $n_0$ such that for all $n \ge n_0$

$$q - \epsilon \lt \frac{a_{n+1}}{a_n} \lt q+\epsilon$$

Multiplying gives us

$$a_{n_0}(q - \epsilon)^{n-n_0} \lt a_{n} \lt (q+\epsilon)^{n-n_0} a_{n_0}$$

and so

$$a_{n_0}^{1/n}(q - \epsilon)^{1-n_0/n} \lt a_{n}^{1/n} \lt (q+\epsilon)^{1-n_0/n} a_{n_0}^{1/n_0}$$

And thus (by taking the limit as $n \to \infty$)

$$ q - \epsilon \le \liminf (a_n)^{1/n} \le \limsup (a_n)^{1/n} \le q + \epsilon $$

Since $\epsilon$ was arbitrary, we have that $q = \lim a_n^{1/n}$.

The case $q=0$, we replace left hand side by $0$, and the proof carries through.

Proof Let us define new sequence $(b_n)_{n=1}^{\infty}$ with $b_1=a_1$ and $b_n=\frac{a_{n+1}}{a_n}$ for all $n>1$.

Now, we consider the following as known,

If $(x_n)_{n=1}^{\infty}$ is a sequence of positive numbers and $\lim x_n=L$, then,