Inspired by a Brilliant problem

$\begin{aligned} \displaystyle n & = \frac{1}{2^1} + \frac{1}{2^3} + \frac{1}{2^5} + \cdots \\ & = \sum_{n \ = \ 1}^{\infty} \frac{1}{2^{2n - 1}} \\ & = \sum_{n \ = \ 1}^{\infty} 2 \cdot \bigg(\frac 14\bigg)^n \\ & = \frac{\displaystyle \frac 12}{\displaystyle 1 - \frac 14} \\ & = \frac 23 \end{aligned}$

Alternative solution:

$\displaystyle \begin{aligned} n & = \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \cdots \\ & = \frac 12 + \frac 14 \left(\frac 12 + \frac 18 + \cdots\right) \\ & = \frac 12 + \frac 14n \\ \frac 34n & = \frac 12 \\ n & = \frac 23 \end{aligned}$

$\frac{1}{2^1}+\frac{1}{2^3}+\frac{1}{2^5}+\dots=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots-(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\dots)=1-\frac{1}{3}=\frac{2}{3}$

This can be easily verified with these figures:

$\large n = \frac{1}{2^1}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\cdots \\ \large \implies 4n = \frac{4}{2^1}+\frac{4}{2^3}+\frac{4}{2^5}+\frac{4}{2^7}+\cdots = 2 + n \\ \large \therefore n = \frac{2}{3}$