Find $\lim_{n \to \infty}\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$

Here is a solution with elementary math.

Let's denote $$S_n = \frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$$

You have

$$2S_n = 1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}$$

Then

\begin{align} S_n = 2S_n - S_n &= 1+\left(\frac{3}{2}-\frac{1}{2} \right)+\ldots+\left(\frac{2n-1}{2^{n-1}} - \frac{2n-3}{2^{n-1}} \right) - \frac{2n-1}{2^{n}} \\ &= 1+\left(1+\frac{1}{2}+\frac{1}{2^2}\ldots+ \frac{1}{2^{n-2}} \right) - \frac{2n-1}{2^{n}} \\ &= 1+2\left(1- \frac{1}{2^{n-1}} \right) - \frac{2n-1}{2^{n}} \\ &= 3-\frac{1}{2^{n-2}} - \frac{2n-1}{2^{n}} \\ \end{align}

Hence, $$\lim_{n\to\infty} S_n=3$$