Consider the family of sequences of the form $.012\ldots n$ for any natural number $n$. So, the sequences in this family are: $.01, .012, .0123, .01234,$ etc. 


Now consider to manipulate each sequence in this way: 


1) Start from the rightmost digit; let's say that the rightmost digit is $n$;  

2) If $n$ is even, replace $n$ by $0$ and add $n/2$ to the digit immediately on the left;  

3) if $n$ is odd, replace $n$ by $1$ and add $(n-1)/2$ to the digit immediately on the left;  

4) keep on modifying the sequence on the left until the only digits are '$0$' and '$1$'.


E.g., suppose that you consider the sequence $.01234.$ Then you modify it in this way: $.01250; .01410; .03010; .11010$.


I want to show that, for any $n$, after rewriting the sequence with only '$0$' and '$1$', I will never obtain any sequence that starts in this way $x.yy$'$y$'... where $x\neq 0$. So, e.g., I will never obtain any sequence like $1.0101$. 


This problem is probably analogous to show that any sequence of the form $(0\*2^{-1})+(1\*2^{-2})+(2\*2^{-3})+...+(n\*2^{-(n+1)})$ is $<1$. However, I am not sure how to prove it.



Decimal binary sequences that cannot be greater than 1
The decimal representation is a red herring. You have an operation that acts on sequences $(a\_0,a\_1,a\_2,\dots)$ by modifying adjacent entries
$$a'\_n=a\_n \pmod{2}$$ 
$$a'\_{n-1}=a\_{n-1}+\lfloor n/2 \rfloor$$
The quantity $\sum \frac{a\_n}{2^n}$ is left invariant at every step and is easily seen to be less than $2$ for any of your finite sequences. This implies the desired conclusion.


