To show limit of $\left|\frac{ a_{n+1}}{a_n}\right|$ is smaller than lim inf $\left|(a_n)^{1/n}\right|$ Suppose that $a_{n}$ is a sequence of real numbers such that $\lim\limits_{n\to\infty} {\frac{\left|a_{n+1}\right|}{\left|(a_n)\right|}}$ exists, 
then $$\lim\limits_{n\to\infty} {\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}} \leq \liminf\limits_{n\to\infty} \left|a_n\right|^{1/n}$$.
This is how I have started:
Suppose  $\lim\limits_{n\to\infty} {\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}}=L, L \in R$
Given any  $\epsilon \gt 0$, there exist $n_0$ such that for all $n \ge n_0$, $$L - \epsilon \lt {\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}} \lt L+\epsilon$$
and after this I am quite clueless how to continue. Any help or hint is much appreciated. Please help if you could. Thank you very much. 
 A: First note that if $\lim\left|{a_{n+1} \over a_n}\right| = L$, then $\liminf\left|{a_{n+1} \over a_n}\right| = L$. Let $A:= \{t : t < \liminf\left|{a_{n+1} \over a_n}\right|\}$. Given any $t \in A$, if we let $\epsilon := L - t$, then there exists an $N$ such that for all $n \geq N$, $L - \epsilon < \left|{a_{n+1} \over a_n}\right| < L + \epsilon$. Or $t < \left|{a_{n+1} \over a_n}\right|$ which is equivalent to $$|a_n| t < |a_{n+1}| \tag{1}.$$ 
Next, we will prove by induction that given any natural number $k$, $|a_n| t^k < |a_{n+k}|$. For the base case, note that the limit inequality holds for all $n \geq N$. Thus, $|a_{n+1}|t < |a_{n+2}|$. Note that if we multiply (1) by $t$ we get $|a_n |t^2 < |a_{n+1}|t < |a_{n+2}|$ proving the base case.
Second, if we assume $|a_n| t^k < |a_{n+k}|$ is true, multiplying by $t$ we get $|a_n |t^{k+1} < |a_{n+k}|t$. Similarly, by the limit inequality, $|a_{n+k}|t < |a_{n+k+1}|$. Thus, the statement is true by induction.
Next, we can take the $(n+k)^{th}$ root of both sides: $|a_n t^k|^{1/(n+k)} < |a_{n+k}|^{1/(n+k)} \implies |a_n|^{1/k}t^{k/(k+n)} < |a_{n+k}|^{1/(n+k)}$.
On the left hand side of the inequality, $$\lim_{k \to \infty} |a_n|^{1/k}t^{k/(k+n)} = t.$$ Thus, the limit infimum also equals $t$. 
Therefore, since the inequality of sequences are preserved in the limits of the sequences, ( http://www.proofwiki.org/wiki/Inequality_of_Sequences_Preserved_in_Limit)
if we take the limit infimum of both sides we get that $$t \leq \liminf_{n \to \infty}|a_{n}|^{1/n}.$$ Since our choice of $t$ was arbitrary, this inequality must hold for all $t \in A$. Thus, $\liminf_{n \to \infty}|a_{n}|^{1/n}$ is an upper bound for the set $A$ such that $$\liminf_{n \to \infty} \left|{a_{n+1} \over a_n}\right| = \lim_{n \to \infty} \left|{a_{n+1} \over a_n}\right| \leq \liminf_{n \to \infty}|a_{n}|^{1/n}$$ by the property of the least upper bound.

P.S. The argument with the set $A$ can be reformulated using this theorem: $$\forall \epsilon > 0 (x - \epsilon < y) \implies x < y$$ I changed this to the equivalent formulation of sets and upper bounds to avoid managing multiple $\epsilon$.
