# Calculation of the Box Dimension

I am writing a presentation on the box dimension and this is an example I've been told to use to illustrate a calculation.

The Exampe: What is the box dimension of $A=\{0,1,\frac{1}{4},\frac{1}{9},\frac{1}{16},\ldots\}$?

My 'attempt': At 'stage' $k$ of the construction, there are $k+1$ points (incuding $0$) with at least $\frac{1}{k^{2}}-\frac{1}{(k+1)^{2}}=\frac{2k+1}{k^{2}(k+1)^{2}}$ between each point. So we take $N_{\delta}(A)$ to be the smallest number of closed balls of radius $\delta$ with $0<\delta<\frac{1}{4}$. Let $k\in\mathbb{Z}$ such that $\frac{1}{k^{2}(k-1)^{2}}>\delta\geqslant\frac{1}{k^{2}(k+1)^{2}}$. If $|U|\leqslant\delta$, then $U$ can cover at most one of the points $\{1,\frac{1}{4},\frac{1}{9},\ldots,\frac{1}{k^{2}}\}$ since $\frac{2k+1}{k^{2}(n+1)^{2}}>\delta$. So at least $n$ sets of diameter $\delta$ are needed to cover $A$, so $N_{\delta}(A)\geq k$, and so $$\frac{\ln(N_{\delta}(A))}{-\ln(\delta)}\geqslant\frac{\ln(k)}{\ln(k^{2}(k+1)^{2})}.$$ Hence $$\underline{\dim}_{box}\geqslant\lim_{\delta\to0}\frac{\ln(k)}{\ln(k^{2}(k+1)^{2})}=\frac{1}{4}.$$

On the other hand, if $\frac{1}{4}>\delta>0$, take $k$ such that $\frac{1}{k^{2}(k-1)^{2}}>\delta\geqslant\frac{1}{k^{2}(k+1)^{2}}$. Then $k+1$ intervals of length $\delta$ can cover $[0,\frac{1}{k^{2}}]$, leaving $k-1$ points of $A$ which can be covered by another $k-1$ intervals. So $N_{\delta}(A)\leqslant2k$, so $$\frac{\ln(N_{\delta}(A))}{-\ln(\delta)}\leqslant\frac{\ln(2k)}{\ln(k^{2}(k-1)^{2})},$$ which gives that $$\overline{\dim}_{box}(A)\leqslant\frac{1}{4}.$$

My problem: I have looked at and used a similar example where $A=\{0,1,\frac{1}{2},\frac{1}{4},\ldots\}$. That example is how I got the inequalities for $\delta$, but I'm not really sure where I plucked them from. Is someone able to help?

Is what I have done correct? - If not, what do I do?