My initial approach is diving the whole sum by $9$ and taking the common $5$ out which gives $\frac{5}{9}[(10-1)+(10-0.1)+(10-0.01)+\cdots + (10-10^{-19})]$ after some algebra this could be reduced to $\frac{5}{9} \times [200-\frac{10}{9} \times (1-10^{-20})]$
after this I am not sure how to show that is almost equal to $110.5$? Also if any body wants to suggest any other tricky/fast way I will appreciate it.