Set with Hausdorff dimension $s = \log 2 / \log 3$ but $H^s = \infty$?

Question

I am trying to solve the following problem:

Find a set $X \subset \Bbb R$ s t.$\dim_H (X)= s$ where $s = \frac{\log 2}{\log 3}$, but $H^s (X) = \infty$.

Here I am using the notations from Fractal Geometry by Kenneth Falconer. From exercise 4.9 of the book. I think I should be looking at some sort of intersection of middle $\lambda$-Cantor sets with $\lambda$ getting closer to $\frac13$ from below. However, I am not sure where to go from there.

Edit: after everyone's suggestions here is what I came up with. Let $C_n$ be a rescale of the middle third Cantor set (call it $C$) which has been translated to fit in $[ \frac{1}{2n}, \frac{1}{2n-1}]$. Let $X = \cup_{n=1}^{\infty} C_n$.

Note $C_n$ and $C_{n+1}$ are $\frac{1}{2n} - \frac{1}{2n+1} = \frac{1}{(2n+1)2n}$ apart. Fix $\varepsilon > 0$ and find $N = N_{\varepsilon}$ such that \begin{equation} \frac{1}{(2N+3)(2N+2)} \leq \varepsilon < \frac{1}{(2N+1)2N}. \end{equation} It follows that \begin{equation} H^s_{\varepsilon} (X) \geq \sum_{n=1}^{N_{\varepsilon}} H^s_{\varepsilon} (C_n) = \sum_{n=1}^{N_{\varepsilon}} H^s_{2n(2n-1)\varepsilon} (C). \end{equation} The rest is clever algebra to show that the above sum diverges to $\infty$ using the fact that $H^s (C) > 0$.