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?

Here's different flavor that's more elementary and explicit (maybe). 
At the $k$-th term among summing $k=1$ to $k = n$, we have the multiplicative factor you got:
\begin{align*}
\frac{ 111\cdots11 }{ 100\cdots00} &= 1 + \frac1{10} + \frac1{100} + \cdots \frac1{10^{k-1}} \tag*{, totally $k$ terms}\\
&= \frac{ 1 - (1/10)^k}{1 - 1/10} \\
&= \frac{10}9 \left( 1 - \frac1{10^k}\right)
\end{align*}
Thus the original summation of $n$ terms is
\begin{align*}
S = \frac3{10}\frac{10}9 \sum_{k = 1}^n \left( 1 - \frac1{10^k}\right) &= \frac13 \cdot \left(n - \frac{ \frac1{10} - \frac1{10^{n+1}} }{1 - \frac1{10}} \right) \\
&=\frac13 \cdot \left(n - \frac19 \left(1 - \frac1{10^n} \right) \right) \\
&= \frac13 \cdot \left(n - \frac19 \frac{ \,\,\overbrace{ 999\cdots 99 }^{n~\text{digits} }}{ \underbrace{ 100\cdots 000 }_{n+1~\text{digits} } } \right) \\
&= \frac13 \left( n -  \frac{ \,\,\overbrace{ 111\cdots 11 }^{n~\text{digits} }}{ \underbrace{ 100\cdots 000 }_{n+1~\text{digits} } }\right) = \frac13 \cdot (n - 0.111\cdots 11)
\end{align*}
There are $n$ digits of $1$ following the decimal point. Or if you prefer,$$S = \frac13 \bigl(n - 1 + 0.888\cdots 89 \bigr)$$where there are $n-1$ digits of $8$ followed by a single $9$.
