Writing 1/3 as a sum of other numbers Is it possible to write $0.3333(3)=\frac{1}{3}$ as sum of $\frac{1}{4} + \cdots + \cdots\frac{1}{512} + \cdots$ so that denominator is a power of $2$ and always different?
As far as I can prove, it should be an infinite series, but I can be wrong.
In case if it can't be written using pluses only, minuses are allowed as well.
For example, $\frac{1}{2}$ is $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$
So, what about $\frac{1}{3}$?

There is no way to write it as a finite sum. For if you bring such a sum to a common denominator, that denominator will be a power of $2$. Minus signs won't help.
It can be expressed as the infinite "sum"
$$\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+\cdots.$$
For note that if $|r|\lt 1$ the geometric series $a+ar+ar^2+ar^3+\cdots$ has sum $\frac{a}{1-r}$. Put $a=\frac{1}{4}$ and $r=\frac{1}{4}$, and simplify.
Another interesting representation of $\frac{1}{3}$ is $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\cdots$.
