Sigma notation for sum of $\ln(x)^2$ from $2$ to $20$ with steps of $0.5$

In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like $$\sum_{s\in S}f(s).$$ In particular you might have $S=\{2,2.5,3,\dots,20\}$.

Does $$\sum_{n = 0}^{36}\ln\left(2+\dfrac{1}{2}n\right)^2$$ satisfy your requirements?

You could change $\ln(x)^2$ into $\ln\left(\frac{x}{2}\right)^2$ to achieve the steps of $0.5$ in this case. You want $\frac{x}{2}$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$\sum_\limits{x = 4}^{40}\ln\left(\frac{x}{2}\right)^2$.