The first addend, $ 1.11111111\ldots $, is by itself equal to the sum of the infinite geometric series $$1+\frac 1{10}+\left(\frac 1{10}\right)^2+\left(\frac 1{10}\right)^3+\cdots,$$which is $ \frac 1{1-\frac 1{10}}=\frac{10}9 $. The second addend is one-tenth of this, so equals $ \frac 19 $. The third addend is one-tenth of the second addend, and so on. Thus the sum of the infinite column of infinite decimals is \begin{align*} \frac{10}9\cdot\left[1+\frac 1{10}+\left(\frac 1{10}\right)^2+\cdots\right] &= \frac{10}9\cdot\frac{10}9 \\ &= \boxed{\frac{100}{81}}.\end{align*}Notice that we have just added $ 1+\frac 2{10}+\frac 3{100}+\frac 4{1000}+\cdots $, sneakily.