If by "lower bound" you mean the minimum, then yes, it is enough to show that for one $n$ you have the period $1$: $\frac{1}{3}=0.33333\ldots$ would suffice.
If by "lower bound" you mean it is the minimum of all accumulation points, i.e. if you need to prove that the period $1$ is reached for infinitely many $n$'s, then it can be done as well. For example:
$$\frac{1}{3}=0.333333\ldots$$
$$\frac{1}{30}=0.033333\ldots$$
$$\frac{1}{300}=0.003333\ldots$$
$$\frac{1}{3000}=0.000333\ldots$$
etc.
(That all excluding "trivial" cases which has a "tail" of zeros, e.g. $\frac{1}{2}=0.500000\ldots$. Not sure how those are regarded in this problem - if you include those, then you are getting an even easier example.)