Is it 0.1234...?

The sum can be expressed as $\dfrac{1}{9}+\dfrac{2}{90}+\dfrac{3}{900}+\cdots$ , but also as $\dfrac{1}{9}\left(1\times\left(\dfrac{1}{10}\right)^0+2\times\left(\dfrac{1}{10}\right)^1+3\times\left(\dfrac{1}{10}\right)^2+\cdots\right)$

In sigma notation this is: $\dfrac{1}{9}\displaystyle \sum_{n=1}^{\infty} n\left(\dfrac{1}{10}\right)^{n-1}$ . Finally, we use the formula for an Arithmetic-Geometric sum:

$\displaystyle \sum_{n=1}^{\infty} nx^{n-1}=\dfrac{1}{(1-x)^2} \\ \dfrac{1}{9}\displaystyle \sum_{n=1}^{\infty} n\left(\dfrac{1}{10}\right)^{n-1}=\dfrac{1}{9} \times \dfrac{1}{(1-\frac{1}{10})^2}=\dfrac{100}{729}=\left(\dfrac{10}{27}\right)^2$

Comparing we get $a=10$ and $b=27$ , hence the answer is $10+27=\boxed{37}$ .

For the sake of variety, I am adding another solution.

As Alan Enrique has mentioned, the summation $S$ can be written as

$S = \dfrac{1}{9} + \dfrac{2}{90} + \dfrac{3}{900} + \ldots - (1)$

$\dfrac{S}{10} = 0 + \dfrac{1}{90} + \dfrac{2}{900} + \ldots - (2)$

Subtracting $(2)$ from $(1)$ ,

$\dfrac{9S}{10} = \dfrac{1}{9} + \dfrac{1}{90} + \dfrac{1}{900} + \ldots$

$\dfrac{9S}{10} = \dfrac{\frac{1}{9}}{1 - \frac{1}{10}} = \dfrac{10}{9^2}$

$S = \dfrac{10^2}{9^3} = \left(\dfrac{10}{27}\right)^2$

$\Rightarrow a = 10 , b = 27 \Rightarrow a+ b = \boxed{37}$