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?

The recursion $10S_n=S_{n-1}+3n$ with $S_0=0$ implies
$$S_n=a\left(\left(1\over10\right)^n-1\right)+bn$$
for some $a$ and $b$.  From $S_1={3\over10}$ and $S_2={63\over100}$ we have
$$\begin{align}
{3\over10}&=a\left({1\over10}-1\right)+b\\
{63\over100}&=a\left({1\over100}-1\right)+2b
\end{align}$$
or
$$\begin{align}
9a-10b&=-3\\
99a-200b&=-63
\end{align}$$
which solves to $b=1/3$ and $a=1/27$, so
$$S_n={1\over27}\left(\left(1\over10\right)^n-1\right)+{1\over3}n$$
