Partition power series $\sum \beta^i=\frac1{1-\beta}$ into three subseries, each with sum $\frac13\frac1{1-\beta}$ Let $\beta \in (0,1)$ fixed, then $\sum\limits_{i=0}^\infty \beta^i =\frac{1}{1-\beta}$.
Can one partition $\{\beta^i\}_{i=0}^\infty$ into, say, $$\{\beta^{i_s}\}_{s=0}^\infty,~\{\beta^{i_l}\}_{l=0}^\infty$, ~\text{and}~ \{\beta^{i_k}\}_{k=0}^\infty$$
such that 
$$\sum_l \beta^{i_l}=\sum_k \beta^{i_k} =\sum_s \beta^{i_s} =\frac{1}{3}\frac{1}{1-\beta} $$
I think it is not possible, although I cannot show it.
If it is possible at which value of $\beta$ will it be possible?
 A: Since $\beta^0=1$, it’s clear that we must have $\frac1{3(1-\beta)}\ge 1$, i.e., $\beta\ge\frac23$, so assume this to be the case. Now follow Did’s suggestion in the comments. Set $I_0=J_0=K_0=\varnothing$. Suppose that $n\in\Bbb N$, and disjoint sets $I_n,J_n$, and $K_n$ have been defined in such a way that 
$$\begin{align*}
&\{k\in\Bbb N:k<n\}\subseteq I_n\cup J_n\cup K_n\;,\\
&\sum_{r\in I_n}\beta^r\le\frac1{3(1-\beta)}\;,\\
&\sum_{r\in J_n}\beta^r\le\frac1{3(1-\beta)}\;,\text{ and}\\
&\sum_{r\in K_n}\beta^r\le\frac1{3(1-\beta)}\;.
\end{align*}\tag{1}$$
Then
$$\begin{align*}
\left(\frac1{3(1-\beta)}-\sum_{r\in I_n}\beta^r\right)&+\left(\frac1{3(1-\beta)}-\sum_{r\in J_n}\beta^r\right)+\left(\frac1{3(1-\beta)}-\sum_{r\in K_n}\beta^r\right)\tag{1}\\
&=\frac1{1-\beta}-\sum_{0\le r<n}\beta^r\\
&=\frac1{1-\beta}-\frac{1-\beta^n}{1-\beta}\\
&=\frac{\beta^n}{1-\beta}\\
&\ge 3\beta^n\;,
\end{align*}$$
since $1-\beta\le\frac13$. This implies that at least one of the three terms in parentheses in line $(2)$ is greater than or equal to $\beta^n$. 


*

*If $\frac1{3(1-\beta)}-\sum_{r\in I_n}\beta^r\ge\beta^n$, let $I_{n+1}=I_n\cup\{n\}$, $J_{n+1}=J_n$, and $K_{n+1}=K_n$. 

*If $\frac1{3(1-\beta)}-\sum_{r\in I_n}\beta^r\beta^n$ and $\frac1{3(1-\beta)}-\sum_{r\in J_n}\beta^r\ge\beta^n$, let $I_{n+1}=I_n$, $J_{n+1}=J_n\cup\{n\}$, and $K_{n+1}=K_n$.  

*If $\frac1{3(1-\beta)}-\sum_{r\in I_n}\beta^r<\beta^n$ and $\frac1{3(1-\beta)}-\sum_{r\in J_n}\beta^r<\beta^n$, let $I_{n+1}=I_n$, $J_{n+1}=J_n$, and $K_{n+1}=K_n\cup\{n\}$.
Then $(1)$ holds with $n$ replaced by $n+1$, and the recursive construction of the sets $I_n,J_n$, and $K_n$ can continue.
Let $I=\bigcup_{n\in\Bbb N}I_n$, $J=\bigcup_{n\in\Bbb N}J_n$, and $K=\bigcup_{n\in\Bbb N}K_n$; it’s straightforward to verify that $\{I,J,K\}$ is a partition of $\Bbb N$ such that
$$\sum_{r\in I}\beta^r=\sum_{r\in J}\beta^r=\sum_{r\in K}\beta^r=\frac1{3(1-\beta)}\;.$$
