You can write your series as $$f(x)=\sum_{1}^{x}\frac{81\times10^k}{(10^k-1)(10^{k+1}-1)}=9\sum_{1}^{x}\frac{1}{10^k-1}-\frac{1}{10^{k+1}-1}$$ Where $x=9$. By telescopy we can show: $$f(x)=\frac{10^{x+1}-10}{10^{x+1}-1}=1-\frac{9}{10^{x+1}-1}$$ So your sum is $S=1-1111111111^{-1}=0.\overline{9999999990}$. So since $2009\equiv 9\pmod{10}$, the digit is $9$.