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 $\frac{1}{3}$?
 A: 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 $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\cdots$.
A: Archimedes showed that if you have a finite sum in which each term is $1/4$ of the previous term, except that the last term is $1/3$ of the previous term, then the sum does not depend on the number of terms, but is just $4/3$ of the first term.  In modern terminology:
$$
1+\frac 1 4 + \frac{1}{16} + \cdots + \frac{1}{4^n} + \frac{1}{4^n}\cdot\frac 1 3\  =\  \frac 4 3.
$$
He deduced from that that the infinite sum
$$
1 + \frac 1 4 + \frac{1}{16} + \cdots = \frac 4 3.
$$
(If the first term is $1/4$ then $4/3$ of that is of course $1/3$.)
A: More generally we can do this for any fraction $\frac{m}{n}$. We know that the sum of the infinite geometric series
$$ S=a+ar+ar^2+ar^3+\dots=\frac{a}{1-r} $$
if $ |r| < 1  $. So we could take $r=-\frac{1}{n-1}$ and $a=\frac{m}{n-1}$ giving
$$ S=\frac{\frac{m}{n-1}}{1+\frac{1}{n-1}}=\frac{\frac{m}{n-1}}{\frac{n}{n-1}}=\frac{m}{n}  $$
We can in fact be very general with both $r$ and $a$. Just suppose $r=\frac{s}{t}$ (with $|s|>|t|$ to ensure $|r|<1$), and let $a=1$ for now then
$$ S=\frac{1}{1-\frac{s}{t}}=\frac{t}{t-s} $$
So if instead we take $a=\frac{m(t-s)}{nt}$ we have again that $S=\frac{m}{n}$.
