Box dimension of $\{\frac{ 1}{5^n} : n \in \mathbb{N}\}$ I am working through a first course in Fractal Geometry, and have encountered a problem that has asked me to calculate the box-counting dimension of $F=\{ \frac{1}{5^n} : n \in \mathbb{N}\}$
However, I am stuck straight away.  Thus far, I've only encountered problems that have asked me to calculate the box-counting dimension of things like the Van Koch curve or variants of it.  So I'm struggling on how to approach this since we only have a set of points, as opposed to a continuous line.
If anyone could perhaps give me a nudge in how I should tackle this, it would be much appreciated. :)

For a sequence decreasing exponentially to zero like this, you might expect the dimension to be zero. To prove it, you need an upper bound on $N_{\varepsilon}(F)$, the number of intervals of length $\varepsilon$ required to cover $F$, for then you can compute
$$\dim(F) = \lim_{\varepsilon\searrow0}\frac{\log(N_{\varepsilon}(F))}{\log(1/\varepsilon)}.$$
To this end, let $\varepsilon > 0$ and let $n$ be the unique positive integer such that
$$\frac{1}{5^{n+1}} \leq \varepsilon < \frac{1}{5^n}.$$
Since $\varepsilon \geq 1/5^{n+1}$, the interval $[0,\varepsilon]$ covers all but $n$ of the points in $F$. Thus $N_{\varepsilon}(F) \leq n+1$. Also, since $\varepsilon < 1/5^n$, 
$$\frac{1}{\log(1/\varepsilon)} < \frac{1}{\log(5^n)}.$$
Putting these inequalities together, we get
$$\frac{\log(N_{\varepsilon}(F))}{\log(1/\varepsilon)} \leq
 \frac{\log(n+1)}{n\log(5)} \to 0$$
as $n\to \infty$.
