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.
 A: $$\frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots=\sum_{n=1}^{\infty} \frac{2n-1}{2^n}$$
A: $$\sum_{n=1}^{+\infty}\frac{2n-1}{2^n}=-1+2\sum_{n=1}^{+\infty}\frac{n}{2^n}=-1+2\sum_{n=1}^{+\infty}\sum_{m\geq n}\frac{1}{2^m}=-1+4\sum_{n=1}^{+\infty}\frac{1}{2^n}=-1+4=3.$$
A: Another approach is to use summation by parts. This method works as follows. If 
$$S_N = \sum_{n=1}^{N} a_n b_n$$
then we can define
$$B_n = \sum_{k=1}^{n} b_k$$
Then
$$S_N = a_N B_N - \sum_{n=1}^{N-1} B_n (a_{n+1} - a_n)$$
In this problem, we can set $a_n = 2n-1$ and $b_n = 1/2^n$. Then
$$B_n = \sum_{k=1}^{n} \frac{1}{2^k} = 1 - \frac{1}{2^n}$$
and $a_{n+1} - a_n = 2$ for all $n$. Therefore,
$$\begin{align}S_N &= a_N B_N - \sum_{n=1}^{N-1} B_n (a_{n+1} - a_n) \\
&= (2N-1)\left(1 - \frac{1}{2^N}\right) - \sum_{n=1}^{N-1} 2\left(1 - \frac{1}{2^n}\right) \\
&= 2N - 1 - \frac{2N - 1}{2^N} - 2(N-1) + 2\sum_{n=1}^{N-1} \frac{1}{2^n} \\
&= 1 - \frac{2N - 1}{2^N} + 2\sum_{n=1}^{N-1} \frac{1}{2^n} \\
\end{align}$$
The middle term goes to zero as $N \rightarrow \infty$, and the rightmost term goes to $2$. So in the limit, we get $1 + 2 = 3$.
A: In this answer, I give a pictorial proof that $$\frac14+\frac28+\frac3{16}+\frac{4}{32}+\frac{5}{64}+\cdots=1$$
With your sum after dividing by $2$ we have $$\frac14+\frac38+\frac5{16}+\frac{7}{32}+\frac{9}{64}+\cdots$$ and the same picture can be used if we expand the original $\frac12\times\frac12$ square to the left as well as to the right and upward. The limit will fill up a rectangle that is $\frac32$ unit wide by $1$ unit tall. So we will have $\frac32$ square units of area. Since we had to halve your series to get this picture, your series sums to $3$.
A: We can take this as the $x=\frac12$ case of the series 
$$f(x)=x+3x^2+5x^3+7x^4+\cdots+(2n-1)x^n+\cdots = \sum_{n=1}^\infty (2n-1)x^n$$
We can recognize in this summation the geometric series 
$$S(x)=x+x^2+x^3+\cdots= x+x\left(x+x^2+\cdots\right)=x+x S(x)=\frac{x}{1-x}$$
where in the last equality we have solved the rest of the equation for $S(x)$.
The obvious thing would be to pull this part of $f(x)$ out of the series above, but it'll actually save us some effort if we consider $f(x)-S(x)$:
$$f(x)-S(x)=2x^2+4x^3+6x^4+\cdots = 2x^2\left[1+2x+3x^2+\cdots\right]=2x^2\sum_{n=1}^\infty n x^{n-1}$$
Do the terms of the bracketed series remind you of anything from calculus?
A: $$\begin{align}
\sum_{k=1}^\infty \frac {2k-1}{2^k} & =\frac {1}{2}+\frac {3}{2^2}+\frac {5}{2^3}+\frac {7}{2^4}+\frac {9}{2^5}+\cdots+\frac {2k-1}{2^k}+\cdots 
\\[1ex]
2\sum_{k=1}^\infty \frac {2k-1}{2^k} & = \sum_{k=1}^\infty \frac {2k-1}{2^{k-1}}
\\ &= \sum_{j=0}^\infty \frac {2j+1}{2^j}  
\\ &= 1 + \sum_{j=1}^\infty \frac {2j+1}{2^j}  
\\[1ex]
\sum_{k=1}^\infty \frac {2k-1}{2^k} & = 1 + \sum_{j=1}^\infty \frac {2j+1-(2j-1)}{2^j}
\\ & = 1 + \sum_{j=1}^\infty \frac{1}{2^{j-1}}
\\ & = \mathop{1 + \underbrace{\sum_{k=0}^\infty \frac{1}{2^{k}}}}_{\text{is this term familiar?}}
\\[2ex] & = 1 + 1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + \cdots + \frac 1 {2^k} + \cdots
\end{align}$$
