Is $\lim_{n \rightarrow \infty} a_{n+1}/a_n=L \implies \lim_{n \rightarrow \infty} \sqrt [n] {a_n}=L$ true? If not, is there a counter example? We were told, in recitation class, about a test for sequences convergence (not series)
Which goes as follows:
if $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}=L$ then $\lim_{n \rightarrow \infty} \sqrt [n] {a_n}=L$.
In a previous question I asked: 
This limit: $\lim_{n \rightarrow \infty} \sqrt [n] {nk \choose n}$.
I was told that this fact is not true.
My question is, can anyone think of a counter example for it?
Because, If yes, Then I would like to let my tutor know about it, but, I don't want to doubt him befor I am sure of it.
Thank you!
 A: I assume we're dealing with (strictly) positive $a_n$ here. Then the assertion is true. Let
$$L = \lim_{n\to\infty} \frac{a_{n+1}}{a_n}.$$
Given $\varepsilon > 0$, choose $N\in\mathbb{N}$ so that
$$M := \max \{0,L-\varepsilon\} < \frac{a_{n+1}}{a_n} < L+\varepsilon$$
for all $n \geqslant N$. Then we have, for $n > N$
$$a_N\cdot M^{n-N} < a_n < a_N\cdot (L+\varepsilon)^{n-N},$$
and taking $n$-th roots
$$\sqrt[n]{a_N} \cdot M^{1-N/n} < \sqrt[n]{a_n} < \sqrt[n]{a_N}(L+\varepsilon)^{1-N/n}.$$
The limit of the lower bound is $M$, and the limit of the upper bound is $L+\varepsilon$, so
$$M \leqslant \liminf \sqrt[n]{a_n} \leqslant \limsup \sqrt[n]{a_n} \leqslant L+\varepsilon.$$
Since $\varepsilon$ was arbitrary, we have indeed
$$\lim_{n\to\infty} \sqrt[n]{a_n} = L.$$
A: Note that the root requires $a_n\geq0$. 
Fix $\varepsilon>0$. Then, for $n$ big enough, $$L-\varepsilon<\frac {a_{n+1}}{a_n}<L+\varepsilon. $$ Then
$$
a_{n+1}\leq (L+\varepsilon)a_n\leq\cdots (L+\varepsilon)^{n}a_1.
$$
So
$$
\sqrt[n]{a_n}\leq(L+\varepsilon)(a_1)^{1/n}.
$$
Then $\limsup_n\sqrt[n]{a_n}\leq(L+\varepsilon)$. As $\varepsilon$ was arbitrary, we get $\limsup_n\sqrt[n]{a_n}\leq L$. 
In a similar way, from $a_{n+1}\geq (L-\varepsilon)a_n$ we get $\liminf_n\sqrt[n]{a_n}\geq(L-\varepsilon)$, so $\liminf_n\sqrt[n]{a_n}\leq L$.
The inequalities 
$$
L\leq\liminf_n\sqrt[n]{a_n}\leq\limsup_n\sqrt[n]{a_n}\leq L$$ show that the limit exists and equals $L$.
