Fractal dimensions on the set $\{0, 1,1/2,\cdots, 1/n, \cdots\}$

Let $X=\{0, 1,1/2,\cdots, 1/n\cdots\}.$ Then, we have $\dim_X E=\dim_P X=0, \dim_B X=1/2$ and $\dim_A X=1.$

Now we will prove that $\dim_AX=1.$

Recall that 

\begin{split}
\dim_A X=\inf\Big\{\alpha&\geq 0 \big|\text{there are constants $b,c>0$ satisfying:} \\
&\text{for any $0<r<R<b$, the inequality $N_{r,R}(X)\leq c(\frac{R}{r})^{\alpha}$ holds}\Big\},
\end{split}

where $N_{r,R}(X)$ denotes the smallest number of balls with radii equal to $r$ need to cover any ball with radius equals to $R.$

Take $R_n=1/n, r_n=1/n^2$ with $n\ge 3$.  Clearly,  $B(0,R_n)\cap X=\{0, 1/n, 1/(n+1)\cdots\}$.

Clearly,  $\left(B(0,R_n) \setminus B(0,r_n)\right)\cap X=\{1/(n^2-1), 1/(n^2-2),\cdots, 1/n\}$. For any $i\in \{1/(n^2-1), 1/(n^2-2),\cdots, 1/n\} , $ we have 

$$i-(i-1)> r_n=1/n.$$

Therefore, 

$$N_{r_n,R_n}(X)\ge 1+((n-1)^2-n+1)=n^2-3n+3\ge n.$$

But $R_n/r_n=n.$  So, $N_{r_n,R_n}(X)\le c (\frac{R}{r})^{\alpha}$ cann't hold for any $\alpha<1$, which implies that $\dim_A X=1.$

 Remark: the centers of the balls $B(x,r_n)$ should be  in $X$!