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

Is it possible to use sigma notation for non-integer steps, for example I want to sum $$\ln(x)^2$$ from $$2$$ to $$20$$ with steps of $$0.5$$, is there a way I could write this in sigma notation or some other form of notation.

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 $$S=\{2,2.5,3,\dots,20\}$$.
• You could also index over the set $S$, to get $\sum_{i=0}^nf(s_i)$, where $S=\{s_0,s_1,\ldots,s_n\}$. – Jam Feb 11 at 10:20
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$$\sum_\limits{x = 4}^{40}\ln\left(\frac{x}{2}\right)^2$$.
Just use $$\frac{x}{2}$$ instead of $$x$$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.