Evaluate the sum without using formulae

$\begin{aligned} S & = \frac 19 \left(\frac 1{10} + \frac 1{10^2} + \frac 1{10^3} + \frac 1{10^4} + \cdots \right) \\ & = \frac 1{90}\color{#3D99F6} \left(1 + \frac 1{10} + \frac 1{10^2} + \frac 1{10^3} + \cdots \right) & \small \color{#3D99F6} \text{An infinite geometric progression} \\ & = \frac 1{90} \times \color{#3D99F6} \frac 1{1-\frac 1{10}} \\ & = \frac 1{90} \times \frac {10}9 \\ & = \boxed{\dfrac 1{81}} \end{aligned}$

Alternative method:

$\begin{aligned} S & = \frac 19 \left(\frac 1{10} + \frac 1{10^2} + \frac 1{10^3} + \frac 1{10^4} + \cdots \right) \\ & = \frac 19 \left(0.1 + 0.01 + 0.001 + 0.0001 + \cdots \right) \\ &= \frac 19 \times 0.1111\dots \\ & = \frac 19 \times \frac 19 \\ & = \boxed{\frac 1{81}} \end{aligned}$

We note that 1/10 = .1, 1/100 = .01, 1/1000 = .001, so adding all the terms gives .1111111111111...…… which we recognize as 1/9. Multiplying by 1/9, we get 1/81. Ed Gray

Instead of the geometric progression technique (which I would recommend if you know how to use it), we can use the fact that our numbering system is base 10. $\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \cdots = 0.1 + 0.01 + 0.001 + \cdots$ and this converges to $0. \overline{111}$ . Furthermore, we know that $0. \overline{111}$ is just $\frac{1}{9}$ . So we get $\frac{1}{9} \cdot \frac{1}{9} = \boxed{\frac{1}{81}}$ .

$(1/9)(a(1-r^2)/(1-r))$
With a being 1/10(lim 1->3) and r being 1/10, It equals 1/9. Then 1/9*1/9=1/9^2=1/81.