Expression generating $\left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \frac{33}{100} + \frac{333}{1000}, \dots \right)$ I'm looking for a closed-form expression (in terms of $n$), that will give the sequence
$$
(s_n) = \left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \frac{33}{100} + \frac{333}{1000}, \dots \right).
$$
Can anyone think of one? I made a related post to this question several minutes ago but I realized I was interpreting the sequence wrong.

Let 
$$S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$ 
$$=\underbrace{\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots}_n$$
$$=3\left(\frac {1}{10}+\frac {11}{100}+\frac {111}{1000} + \frac {1111}{10000}+\cdots\right)$$
$$=3\left(\frac {1}{10^1}+\frac {11}{10^2}+\frac {111}{10^3} + \frac {1111}{10^4}+\cdots\right)$$
$$=3\left(\frac {\sum_{k=0}^010^k}{10^1}+\frac {\sum_{k=0}^110^k}{10^2}+\frac {\sum_{k=0}^210^k}{10^3} + \frac {\sum_{k=0}^310^k}{10^4}+\cdots+\frac {\sum_{k=0}^{(n-1)}10^k}{10^n}\right)$$
So $$\frac{1}{10^{n}}\left(\sum_{k=0}^{n-1}10^k\right)= \frac{1}{10^{n}}\left(\frac{10^{n}-1}{9}\right) = \frac{1}{9}\left(1 - \frac{1}{10^n}\right) = \frac{1}{9}\left(1 - 10^{-n}\right)$$
Therefore
\begin{align}
S &= 3\left(\sum_{k=1}^n\left[\frac{1}{9}\left(1 - 10^{-k}\right)\right]\right)\\
&= \frac{3}{9}\sum_{k=1}^n\left(1 - 10^{-k}\right)\\
&= \frac{1}{3} \left( \sum_{k=1}^n 1 - \sum_{k=1}^n10^{-k}\right)\\
&= \frac{1}{3} \left( n- \sum_{k=1}^n10^{-k}\right)\\
&= \frac{1}{3} \left( n- \frac{1}{9}\left(1 - 10^{-n}\right)\right)
\end{align}


$$S = \frac{1}{27}\left(10^{-n}+ 9n -1\right) $$

