# Hausdorff dimension of Sierpinski triangle less than log3/log2

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, $$m(α)=\lim_{\delta\to 0^+} H_β^δ (S)=0$$