= Solution
First suppose $L>0$. Taking logarithms,
$$
\log a_n
=\log a_1+\sum_{k=1}^{n-1}
\log\left(\frac{a_{k+1}}{a_k}\right).
$$
Divide by $n$. Since the summands tend to $\log L$, their Cesàro means also tend to $\log L$. Hence
$$
\frac1n\log a_n\to\log L,
$$
and exponentiation gives $a_n^{1/n}\to L$.
If $L=0$, then for every $\varepsilon>0$ the ratios are eventually at most $\varepsilon$, so $a_n\leq C\varepsilon^n$ for a suitable constant $C$. Thus
$$
\limsup a_n^{1/n}\leq\varepsilon.
$$
Letting $\varepsilon\downarrow0$ proves the <ratio limit implies root limit> in all cases.
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