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.

I saw this proof today and thought it's a nice one:
Let  $a_n\ge 0$, $\lim\limits_{n \to \infty}a_n=L$. So there are 2 options:
(1) $L>0$:
$$
\lim\limits_{n \to \infty}a_n=L
\iff \lim\limits_{n \to \infty}\frac{1}{a_n}=\frac{1}{L}$$
Using Cauchy's Inequality Of Arithmetic And Geometric Means we get:
$$\frac{n}{a_1^{-1}+\dots+a_n^{-1}}\le\sqrt[n]{a_1\cdots a_n}\le \frac{a_1+ \cdots + a_n}{n}$$
Applying Cesaro Theorem on $a_n$, notice that RHS $\mathop{\to_{n \to \infty}} L$ , and that by applying Cesaro Thm on $1/a_n$, LHS$\mathop{\to_{n \to \infty}} \frac{1}{1/L}=L$ . And so from the squeeze thm:
$$\lim\limits_{n \to \infty}\sqrt[n]{a_1\cdots a_n}=L$$
(2) $L=0$:
$$0\le\sqrt[n]{a_1\cdots a_n}\le \frac{a_1+ \cdots + a_n}{n} $$
$$\Longrightarrow\lim\limits_{n \to \infty}\sqrt[n]{a_1\cdots a_n}=0=L$$
Now, define:
$$b_n = \begin{cases}{a_1}  &{n=1}\\\\ {\frac{a_n}{a_{n-1}}} &{n>1}\end{cases}$$
and assume $\lim\limits_{n \to \infty}b_n=B $.
Applying the above result on $b_n$ we get:
$$\frac{n}{b_1^{-1}+\dots+b_n^{-1}}\le\sqrt[n]{b_1\cdots b_n}\le \frac{b_1+ \cdots + b_n}{n}$$
$$\frac{n}{b_1^{-1}+\dots+b_n^{-1}}\le\sqrt[n]{a_1\cdot (a_2/a_1)\cdots (a_n/a_{n-1})}\le \frac{b_1+ \cdots + b_n}{n}$$
$$\frac{n}{b_1^{-1}+\dots+b_n^{-1}}\le\sqrt[n]{a_n}\le \frac{b_1+ \cdots + b_n}{n}$$
$$\Longrightarrow\lim\limits_{n \to \infty}\sqrt[n]{a_n}=B$$
So we can conclude that if $\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}$ exists and equal to $B$, then $\lim\limits_{n \to \infty}\sqrt[n]{a_n}=B$.
