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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 $\frac{1}{3}$?

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The answer by André Nicolas could be taken to fully answer the question, so that usually I would not add another answer. But I added one that has a different point of view that I think is worth knowing about. –  Michael Hardy Aug 25 '12 at 13:55

3 Answers 3

up vote 6 down vote accepted

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 $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\cdots$.

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It may not be "finite" but the solution is "closed", the OP should appreciate this. –  Squirtle Aug 24 '12 at 18:01
Wow, thanks for the second solution!!!! One more question then, if it is infinite number of possible ways to write 1/3. I'm thinking as well. BTW, I thought for 2 hours on my question prior to register as a new member. Geometric progression, that was it! ) –  Haradzieniec Aug 24 '12 at 18:07
Another way (but is is repeated, so that's not a solution on my question) is 1/8 +1/8 + 1/32 + 1/32 + 1/128+ 1/128+.... –  Haradzieniec Aug 24 '12 at 18:11
@Haradzieniec: You are right, there is an infinite number of representations, even with the proviso that nothing can be repeated. –  André Nicolas Aug 24 '12 at 18:14

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 $1/3$.)

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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 $S=\frac{m}{n}$.

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