Sum of an unorthodox infinite series $ \frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots $
This is a pretty unorthodox problem, and I'm not quite sure how to simplify it.  Could I get a solution?  Thanks.

$$\begin{align}
\sum_{k=1}^\infty \frac {2k-1}{2^k} & =\frac {1}{2}+\frac {3}{2^2}+\frac {5}{2^3}+\frac {7}{2^4}+\frac {9}{2^5}+\cdots+\frac {2k-1}{2^k}+\cdots 
\\[1ex]
2\sum_{k=1}^\infty \frac {2k-1}{2^k} & = \sum_{k=1}^\infty \frac {2k-1}{2^{k-1}}
\\ &= \sum_{j=0}^\infty \frac {2j+1}{2^j}  
\\ &= 1 + \sum_{j=1}^\infty \frac {2j+1}{2^j}  
\\[1ex]
\sum_{k=1}^\infty \frac {2k-1}{2^k} & = 1 + \sum_{j=1}^\infty \frac {2j+1-(2j-1)}{2^j}
\\ & = 1 + \sum_{j=1}^\infty \frac{1}{2^{j-1}}
\\ & = \mathop{1 + \underbrace{\sum_{k=0}^\infty \frac{1}{2^{k}}}}_{\text{is this term familiar?}}
\\[2ex] & = 1 + 1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + \cdots + \frac 1 {2^k} + \cdots
\end{align}$$
