Let $f(n)$ denote the corresponding expression with the digit 2 appearing exactly $n$ times. Then $f(1)=\frac{1}{2}$ and for $n>1, f(n)=\frac{1}{2-f(n-1)}$. By induction using the identity $\frac{1}{2-\frac{N-1}{N}}=\frac{N}{N+1}$, $f(n)=\frac{n}{n+1}$ for all $n \geq 1$, so $f(2013)=\frac{2013}{2014}$.