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!
 A: Each step of your formula adds another division by two, this is the same as multiplying with $\frac12$. If you want to multiply a number with something a set number of times, you use exponents. This gives a formula for each component in the sum.
$$
x\cdot\left(\frac12\right)^n
$$
This means that the sigma notation of your sum would be:
$$
\sum_{n=1}^\infty x\cdot\left(\frac12\right)^n=x\sum_{n=1}^\infty\left(\frac12\right)^n
$$
The value of that infinite sum is $1$, so it turns out that the above expression doesn't evaluate to $2x-1$, instead:
$$
\sum_{n=1}^\infty x\cdot\left(\frac12\right)^n=x
$$

To evaulate the sum, we define $S_N$ to be
$$
S_N=\sum_{n=1}^N\left(\frac12\right)^n=\frac12+\frac14+\cdots+\frac1{2^N}
$$
Notice that the value of $\frac12S_N$ is given as below:
$$
\frac12S_N=\sum_{n=2}^{N+1}\left(\frac12\right)^n=\frac14+\frac18+\cdots+\frac1{2^{N+1}}
$$
Subtracting these from each other, and arranging the sum the following way gives:
$$
S_N-\frac12S_N=\frac12+\frac14-\frac14+\cdots+\frac1{2^N}-\frac1{2^N}-\frac1{2^{N+1}}
$$
Canceling terms on the right side and simplying the left side gives
$$
\frac12S_N=\frac12-\frac1{2^{N+1}}
$$
Multiply by 2 and join the fractions
$$
S_N=\frac{2^N-1}{2^N}=1-2^{-N}
$$
To evaluate the infinte sum, simply take the limit as $N\to\infty$, since $2^{-N}\to0$ as $N\to\infty$ we have the limit as
$$
\sum_{n=1}^\infty\left(\frac12\right)^n=\lim_{N\to\infty}S_N=1
$$
A: 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 $|a|<1$.
