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Hi there I am struggling to understand the Hausdorff dimension of the Sierpinski triangle $S$. Below is I did to prove that $\alpha=\frac{\log 3}{\log 2}$, what should I do for $\alpha \le \frac{\log 3}{\log 2}$?

Given $δ>0$, we choose $k$ so that $2^{-k}<δ$.

Since the set $S_k$ covers $S$ and consists of $3^k$ triangles each of diameter $2^{-k}<δ$, we have $$H_α^δ (S)≤3^k (2^{-k} )^α=\frac{3^k}{2^kα}.$$

By definition of Hausdorff dimension, we know that (1) $∀β>α, m_β (S)=0$ and (2) $∀β<α,m_β (S)=∞$.

If $α<\frac{\log ⁡3}{\log ⁡2}$, then $3^k/2^{kα} ≥3^k/3^k =1$. Hence, $\displaystyle{\lim_{k\to \infty}⁡ \frac{3^k}{2^{kα}}=∞}$; and we have, $$m(α)=\lim_{\delta\to 0^+} H_β^δ (S)=\infty$$

If $α>\frac{\log ⁡3}{\log ⁡2}$, then $3^k/2^{kα} ≤3^k/3^k =1$. Hence, $\displaystyle{\lim_{k\to \infty}⁡ \frac{3^k}{2^{kα}}=0}$; and we have, $$m(α)=\lim_{\delta\to 0^+} H_β^δ (S)=0$$

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