I am writing a proof for my measure theory course that seems to boil down to the existence of a sequence of the form $$ \left\{\frac{1}{2^{n_1}}, \frac{1}{2^{n_2}}, \ldots \right\} $$ with $n_k \in \mathbb{N}, \forall k \in \mathbb{N}$ (wlog we may assume that $n_1 < n_2 < \ldots$) such that $$ \sum_{k=1}^\infty \frac{1}{2^{n_k}} = \frac{3}{8} $$
How can I show that such a sequence either exists or doesn't exist? Moreover, if such a sequence does exist, is it possible to write in closed form; that is, can I define the sequence of powers in closed form (if such a sequence exists but this is not possible, my proof will unfortunately not work R.I.P.)
Preferably, I should be able to write $n_1, n_2, \ldots$ such that the sequence is of the form $$ \left\{\frac{1}{(2^l)^1}, \frac{1}{(2^l)^2}, \frac{1}{(2^l)^3}, \ldots\right\} $$ for some $l \in \mathbb{N}$, as otherwise my approach to the proof again becomes basically impossible, as it involves "removing" the sum of a defined sequence of the form $\{\frac{1}{2^{n_1}}, \frac{1}{2^{n_2}}, \ldots \}$ from $\sum_{n=1}^\infty \frac{1}{2^n}$.
Any hints or suggestions are appreciated!
For $\frac 38$ you can just exhibit it. For an arbitrary desired sum in $(0,1]$ you can use the binary expansion. If the desired sum is a dyadic rational you need to choose the expansion with all $1$s at the end, not the one with all $0$s at the end.