Sum to $n$ terms of the given series Find the sum to $n$ terms of the given series:
$$0.3+0.33+0.333+0.3333+\cdots$$
My Attempt:
Let 
$$S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$ 
$$=\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots$$
$$=\frac {3}{10} \left[1+\frac {11}{10}+\frac {111}{100}+\frac {1111}{1000}+\cdots \text{ to $n$ terms}\right]$$
How do I continue from here?

$$\dfrac {3}{10} \left(1+\dfrac {11}{10}+\dfrac {111}{100}+\dfrac {1111}{1000}+\ldots\right) \\
=\dfrac {3}{10} \left(\dfrac{1}{10^0}+\dfrac {10 +1}{10^1}+\dfrac {10^2 +10 +1}{10^2}+\dfrac {10^3+10^2+10+1}{10^3}+\ldots\right)
$$
Each fraction has a (finite) geometric series in the denominator (there is a formula for those) and a power of ten in the denominator.
$$=\frac3{10}\sum_{k=0}^{n-1} \frac1{10^k} \sum_{j=0}^k 10^j .$$
