If $a_{n+1}/a_n\to\ell$ show $a_n^{1/n}\to\ell$ 
Suppose we have a sequence $(a_n)$ such that $a_n\neq0$ and $$\lim\dfrac{a_{n+1}}{a_n}=\ell$$ Show that $$\lim a_n^{1/n}=\ell.$$

I have a proof sketch, but it needs an extra assumption:
By the definition of convergence $\forall\varepsilon>0\exists N\forall n>N,\ \ell-\varepsilon<\dfrac{a_{n+1}}{a_n}<\ell+\varepsilon$.
Here is where I have to assume that eventually $a_n>0$ to conclude (by induction) that eventually $$(\ell-\varepsilon)^na_0<a_n<(\ell+\varepsilon)^na_0\\\therefore(\ell-\varepsilon)a_0^{1/n}<a_n^{1/n}<(\ell+\varepsilon)a_0^{1/n}\\\therefore \ell-\varepsilon\leq\lim a_n^{1/n}\leq\ell+\varepsilon\\\therefore\lim a_n^{1/n}=\ell$$
But I haven't yet figured out how to prove this without the extra assumption.
 A: Suppose $$a_n >0 \quad\frac{a_{n+1}}{a_n}\to\ell \quad \ell\in [0,+\infty)$$
Thus $\forall\varepsilon>0, \exists n_0\in \mathbb{N}$ s.t. $\forall n\ge n_0$
$$\ell-\frac \varepsilon2\le\frac{a_{n+1}}{a_n}\le\ell+\frac \varepsilon2$$
Therefore
$$\ell-\frac \varepsilon2\le\frac{a_{n_0+1}}{a_{n_0}}\le\ell+\frac \varepsilon2 $$
$$\left(\ell-\frac \varepsilon2 \right)^ka_{n_0}\le a_{n_0+k}\le\left(\ell+\frac \varepsilon2\right)^k a_{n_0}$$
$$\left(\ell-\frac \varepsilon2 \right)^n\left(\ell-\frac \varepsilon2 \right)^{-n_0} a_{n_0}\le a_{n}\le\left(\ell+ \frac \varepsilon2\right)^n \left(\ell+ \frac \varepsilon2\right)^{-n_0} a_{n_0}$$
$$\left(\ell-\frac \varepsilon2 \right)\sqrt[n]{\left(\ell-\frac \varepsilon2 \right)^{-n_0} a_{n_0}}\le \sqrt[n]{a_{n}} \le \left(\ell+ \frac \varepsilon2\right) \sqrt[n]{\left(\ell+ \frac \varepsilon2\right)^{-n_0} a_{n_0}}$$
Thus eventually:
$$\ell-\varepsilon \le \sqrt[n]{a_{n}} \le\ell+ \varepsilon \iff \sqrt[n]{a_{n}}\to \ell$$
Note: 
for $\ell=0$ consider:
$$0\le\frac{a_{n_0+1}}{a_{n_0}}\le\ell+\frac \varepsilon2 $$
A: To avoid confusions, I will assume 


*

*$\color{red}{\ell >0}$

*$a_n >0$ so that $\sqrt[n]{a_n}$ makes sense for all $n$.

*and $\varepsilon$ mentioned below is $\varepsilon< \ell$.


We are given that $\forall\varepsilon>0, \exists N(\varepsilon)\in \mathbb{N}$ s.t. $\forall n> N(\varepsilon)$
$$\ell-\varepsilon<\frac{a_{n+1}}{a_n}<\ell+\varepsilon \tag{1}$$
Now
$$\sqrt[n]{a_n}=\sqrt[n]{\frac{a_{n}}{a_{n-1}}\cdot  \frac{a_{n-1}}{a_{n-2}}\cdot ...\cdot \frac{a_{N(\varepsilon)+2}}{a_{N(\varepsilon)+1}} \cdot a_{N(\varepsilon)+1}}$$
or
$$\sqrt[n]{a_{N(\varepsilon)+1}}\sqrt[n]{\left(\ell-\varepsilon\right)^{n-N(\varepsilon)-1}} < \sqrt[n]{a_n} <\sqrt[n]{a_{N(\varepsilon)+1}} \sqrt[n]{\left(\ell+\varepsilon\right)^{n-N(\varepsilon)-1}} \iff \\
\left(\ell-\varepsilon\right) \cdot \sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell-\varepsilon)^{N(\varepsilon)+1}}} < \sqrt[n]{a_n} <\left(\ell+\varepsilon\right) \cdot \sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell+\varepsilon)^{N(\varepsilon)+1}}} \tag{2}$$
Both
$$\sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell-\varepsilon)^{N(\varepsilon)+1}}} \rightarrow 1, n\rightarrow\infty$$
$$\sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell+\varepsilon)^{N(\varepsilon)+1}}} \rightarrow 1, n\rightarrow\infty$$
and from some $n$ onwards (or maximum between it and $N(\varepsilon)$)
$$\sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell-\varepsilon)^{N(\varepsilon)+1}}} > 1-\frac{\varepsilon}{\ell}$$
$$\sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell+\varepsilon)^{N(\varepsilon)+1}}} < 1+\frac{\varepsilon}{\ell} \tag{3}$$
or
$$\left(\ell-\varepsilon\right) \cdot \left(1-\frac{\varepsilon}{\ell}\right) < \sqrt[n]{a_n} <\left(\ell+\varepsilon\right) \cdot \left(1+\frac{\varepsilon}{\ell}\right)$$
or (because I assumed $\varepsilon< \ell$)
$$\ell - 3\varepsilon<\ell-\varepsilon -\varepsilon + \frac{\varepsilon^2}{\ell} < \sqrt[n]{a_n} <\ell+\varepsilon+\varepsilon+\frac{\varepsilon^2}{\ell} < \ell + 3\varepsilon \tag{4}$$
If we "adjust" $(1)$ above to $\frac{\varepsilon}{3}$ then $(4)$ becomes
$$\ell - \varepsilon< \sqrt[n]{a_n} < \ell + \varepsilon$$

One obvious question is, what happens when $\color{red}{\ell =0}$?
Then $(1)$ becomes
$$0<\frac{a_{n+1}}{a_n}< \varepsilon \tag{1a}$$
Then $(2)$ becomes
$$0<\sqrt[n]{a_n} <\varepsilon \cdot \sqrt[n]{\frac{a_{N(\varepsilon)+1}}{\varepsilon^{N(\varepsilon)+1}}} \tag{2a}$$
Then $(3)$ reduces to
$$\sqrt[n]{\frac{a_{N(\varepsilon)+1}}{\varepsilon^{N(\varepsilon)+1}}} < 1+\varepsilon \tag{3a}$$
And $(4)$ becomes (assuming $\varepsilon <1$)
$$0< \sqrt[n]{a_n} <2\varepsilon \tag{4a}$$
If we "adjust" $(1a)$ above to $\frac{\varepsilon}{2}$ then $(4a)$ becomes
$$0< \sqrt[n]{a_n} < \varepsilon$$
