Given $a_n > 0$, $b_n = \frac{a_{n+1}}{ a_n}$ and $c_n = (a_n) ^{1/ n }$. Show $b_n → L \Rightarrow c_n → L$. Suppose that for every $n ∈ N$, $a_n > 0$ and set $b_n = \frac{a_{n+1}}{ a_n}$ and $c_n = (a_n)^{1/ n }$. Prove that if $b_n → L$ then $c_n → L$. Is the converse true.
If $a_n \to \infty$ then $b_n \to 1$ and similarly $c_n \to 1$
So it's easy to see this is the case. Seems like the squeeze theorem is the trick here. But I know $\neg(c_n \le b_n), \forall a_n$ 
but maybe: $ ??\le (a_n)^{1/n} \le (a_n)^{1/n} \cdot \frac{a_n+1}{a_n} $
 A: We give a semi-formal argument, for positive $L$. A slightly simpler argument works for $L=0$.
Let $\epsilon$ be very small positive, much smaller than $L$. There is an $N$ such that if $n\ge N$ then $L-\epsilon/2\lt \frac{a_{n+1}}{a_n}\lt L+\epsilon/2$.
It follows that for any $m$ we have 
$$a_N(L-\epsilon/2)^{m}\lt a_{N+m}\lt a_N(L+\epsilon/2)^m.$$ 
Take the $(N+m)$-th root. If $m$ is large enough, $a_N^{1/(N+m)}(L-\epsilon/2)^{m/(N+m)}$ is larger than $L-\epsilon$, and $a_N^{1/(N+m)}(L+\epsilon/2)^{m/(N+m)}$ is smaller than $L+\epsilon$. 
So given $\epsilon\gt 0$, for large enough $k$ we have that $a_k^{1/k}$ is within $\epsilon$ of $L$.
Remark: The converse does not hold. Let $a_n=1$ if $n$ is odd, and let $a_n=2$ if $n$ is even.
A: Let $x_n=\log b_n\to \log L$, then 
$$\log c_n=\frac{C+x_1+\cdots+x_n}n$$
for some constant C. We have a direct case of Cesaro mean.
A: Hint:
Use the classical result whose proofs have been asked at least one time here:
If $(a_{n})$ is a sequence of $a_{n} > 0$, then
$$
\liminf \frac{a_{n+1}}{a_{n}} \leq \liminf_{n \to \infty}a_{n}^{1/n}  \leq \limsup a_{n}^{1/n} \leq \limsup \frac{a_{n+1}}{a_{n}}.
$$
From this it follows immediately that if $\lim_{n \to \infty}a_{n+1}/a_{n}$ exists and $=l$ for some $l \in \mathbb{R}$, then $\lim_{n \to \infty}a_{n}^{1/n}$ exists and $=l$.
