Refer to my previous question: Can every positive real number $\leqslant\frac{\pi^2}6$ be expressed in this form?
Let $$A = \left\{\sum_{j\in S}f(j) : S\subset\mathbb N\setminus\{0\} \right\}$$ where $f:\mathbb N\setminus\{0\}\to\mathbb R$ is a sequence such that $\sum_{j=1}^\infty f(j)<\infty$. It was shown in my previous question, where $f(j) = \frac1{j^2}$, that the set $A$ was the disjoint union of compact intervals. Is it ever the case where $A$ is connected?
The motivation is that $X_n\stackrel{\mathrm{i.i.d.}}\sim\mathrm{Ber}(p)$, $Y(n) = Xf(n)$ and $S = \sum_{n=0}^\infty Y_n<\infty$. I would like to see if it is possible to define $f(j)$ such that $S$ is a continuous random variable. My intuition is that this is not possible due to the nature of $X_1$ taking values in $\{0,1\}$ - there is no way to "smooth out" this distribution so that the limiting distribution $S$ is continuous. How can I make this argument rigorous?
