The Soup Problem: how to asymptotically fairly split a geometric series and a constant one using a single pattern?

Proposition. Let $(a_n)$ be an absolutely convergent series of real numbers such that for any natural $k$ we have $$|a_{2k-1}-a_{2k}|\le \sum_{n=k+1}^\infty |a_{2n-1}-a_{2n}|.\label{1}\tag{1}$$ Then there exists a subset $M$ of natural numbers such that $|M\cap \{2k-1,2k\}|=1$ for each natural $k$ and $\sum_{n\in M} a_n=\tfrac 12\sum_{n=1}^\infty a_n$.

Proof. For each natural $k$ put $b_k=|a_{2k-1}-a_{2k}|$. It suffices to show that we can consecutively choose signs for $b_k$, providing $\sum_{k=1}^\infty \pm b_k=0$. At the beginning choose a sign “$+$” for $b_1$ and for each $k>1$ choose for $b_k$ a sign “$+$”, if $\sum_{i=1}^{k-1} \pm b_i\ge 0$ and “$-$”, otherwise. It is easy to check that Condition \eqref{1} provides $\sum_{k=1}^\infty \pm b_k=0$. $\square$

Corollary. For each $q\in (1/\sqrt{2},1)$ there exists a subset $M$ of natural numbers such that $|M\cap \{2k-1,2k\}|=1$ for each natural $k$ and $\sum_{n\in M} q^n=\tfrac 12\sum_{n=1}^\infty q^n$.

Proof. It suffices to check that for any any natural $k$ we have $$q^{2k-1}-q^{2k}\le \sum_{n=k+1}^\infty q^{2n-1}-q^{2n}=\frac{q^{2k+1}}{1+q}. \square$$

I wrote a short paper devoted to your problem.

Solving it, I introduced the following notion, considered in a separate question. A number $q\in (1/2, 1)$ is approximating, if there exist non-negative numbers $A$ and $N$ such that for each $x_0\in [0,A]$ there exist $n\le N$ and a polynomial $P(x)$ of the form $\sum_{i=1}^n \pm x^i$ such that $P(1)=0$ and $|x_0-P(q)|\le Aq^n$.

Proposition. For each approximating number $q$ there exists a subset $M$ of natural numbers such that $\lim_{N\to\infty} |M \cap \{1,\ldots, N\} |/N = 1/2$, and $\sum_{n\in M} q^n=\tfrac 12\sum_{n=1}^\infty q^n$.

Proof. Let the number $q$ is approximating with the constants $A$ and $N$. Put $k_0=0$. Then we can inductively build an increasing sequence $(k_n)$ of natural numbers such that $k_{n+1}-k_n\le N$ for each $n$, assigning signs to numbers $q^i$ for $k_{n-1}<i\le k_n$ (a half of the assigned signs are “$+$” and the other half are “$-$”) assuring $\sum_{i=1}^{k_n} \pm q^i\le Aq^{k_n}$. Let $M$ be the set of natural $n$ such that $q^n$ has “$+$” sign. $\square$

In the paper is shown that each $q\in (q_\infty,1)$ is approximating, where $q_\infty=0.5845751\dots$ is a unique positive root of a polynomial $Q_\infty(x)=x^4+x^3+2x^2-1$.