<s>[INST] 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.
 [/INST] 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 $1 + 2 = 3$.
</s>