Proving sum of negative powers of 2 equals one before becoming greater than one Let $x$: $x_1 \geq x_2 \geq x_3\geq...\geq x_n$ be negative powers of 2 with sum greater than one. Then $\exists$ $l$ s.t. $x_1 + x_2 + ... + x_l = 1$. It seems obvious from examples, but I'm finding it a bit difficult to prove it. Any help would be highly appreciated.

Consider the partial series, up until the value passes $1$. Considering the last value used, we can calculate the minimum remaining number of steps ($MRS_i$) until the series reaches $1$, and this will always be an integer.
We can start with a partial series value of $0$ and the assumption that previous value $x_0 = 1$. This gives us $MRS_0 = 1$.
Now examining the next value, we can recalculate $MRS$ in the light of this. If $x_i<x_{i-1}$, we known that $x_{i-1}$ exactly divides $x_{i}$ and we multiply the previous $MRS$ by $\dfrac{x_i}{x_{i-1}}$, an integer, and either way we then subtract one, to again get an integer $MRS_i$. So as we apporach a partial series value of $1$ it is always an integer number of terms away, until $MRS_\ell = 0$ as required for the full series to pass $1$.
Note that this results holds whenever $x_i \mid x_{i-1}$ is a property of the sequence - we don't actually need the successive $x_i$ to be (negative) powers of two. Also, if the series passes other integers, it will also hit them exactly.
