jmc

We begin this problem by summing $0.\overline{1}$, $0.\overline{01}$, and $0.\overline{0001}$ as decimals. We do this by realizing that $0.\overline{1}$ can also be written as $0.\overline{1111}$ and that $0.\overline{01}$ can be written as $0.\overline{0101}$. Thus, $0.\overline{1}+0.\overline{01}+0.\overline{0001}=0.\overline{1111}+0.\overline{0101}+0.\overline{0001}=0.\overline{1213}$. (Since there is no carrying involved, we can add each decimal place with no problems.)

To express the number $0.\overline{1213}$ as a fraction, we call it $x$ and subtract it from $10000x$: \begin{array}{r r c r@{}l} &10000x &=& 1213&.12131213\ldots \\ - &x &=& 0&.12131213\ldots \\ \hline &9999x &=& 1213 & \end{array} This shows that $0.\overline{1213}=\boxed{\frac{1213}{9999}}$.

(Note: We have to check that this answer is in lowest terms. The prime factorization of $9999$ is $3^2\cdot 11\cdot 101$, so we must check that $1213$ is not divisible by $3$, $11$, or $101$.

Since $1+2+1+3=7$ is not a multiple of $3$, neither is $1213$. Also, $1213=11^2\cdot 10+3=101\cdot 12+1$, so $1213$ can't be a multiple of $11$ or $101$.)