Mathematically Expressing the Sum of... I noticed that $$\large{2x-1 = \frac{x}{2} + \frac{\frac{x}{2}}{2} + \frac{\frac{\frac x2}2}2} + \cdots$$ until the output of one of the steps in the pattern equals $1$. Or in other words $2x-1$ is equal to the sum of $\large{\frac{x}{2}}$ plus that result divided by two plus that result divided by two and so forth.
How do I notate this with a Sigma? Or would I not use Sigma and if so what is the best way to notate this?
I cannot figure it out; any help will be appreciated!

The partial sum of this geometric series is $\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}$, if you start at k=0. Thus in your case it is $\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}-1$, because $2^0=1$. 
$=\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}-\frac{1-\frac{1}{2}}{1-\frac{1}{2}}$=$\frac{\frac{1}{2}-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}$
If $n \to \infty$ then $\lim_{n \to \infty} \frac{\frac{1}{2}-\left( \frac{1}{2} \right) ^n}{\frac{1}{2}}=\frac{\frac{1}{2}-0}{\frac{1}{2}}=1$.
Consider that $\lim_{n \to \infty} a^n=0$, if $|a|<1$.
