How to express series summation notation in reverse? Recently started learning about series summation.
For example
$\Large \frac{1}2+\frac{1}4+\frac{1}8+...=\sum_{n=1}^{\infty} (\frac{1}2)^n=1$
and can be visually proved with square area $1$
containing the infinite series summation from large to small areas:

However, is it possible to express a reverse series summation from small to large areas?
$\Large ...+ \frac{1}{2^n}+\frac{1}{2^{n-1}}+...+\frac{1}2=\sum_{n=\infty}^{1} (\frac{1}2)^n=1$
The idea is to convey meaning by notation. For example, emptying a square by half each step while the reverse is filling it.
Thanks.
 A: So you will have to start somewhere, with a first term, call it a.
Let's say the terms represent area as in the example, then they will be positive, and if they go from smaller to larger then you will expect all of the remaining terms to be bigger than the first term a.
This represents a problem since we expect that the sum of infinitely many terms  bigger than a lower bound a should diverge.
So for instance if you started with a square and kept doubling it, it would diverge.
All of that being said, the phrase 'express a reverse series' which you used is vague enough that it is hard to definitively answer no to the question.
If an expression such as the one you gave in the statement meets the criteria that you set out, then that would suggest that it can be done for any series, since you are just re-arranging a finite segment of the series to that the smallest are ordered first and having the usual trailing dots go from right to left instead of left to right and be located in the front.
We can surely rearrange how we write series so that the smallest terms are left most and the trailing dots go from right to left and are located in front just as you demonstrated there. The notation once it is defined and agreed upon is unambiguous.
The phrase 'small to large areas' indicates to me that the first interpretation is more along the lines of what you are thinking, in that case then no. If those areas are supposed to represent the partial sums of the sequence then they will move from small to large and the answer is yes.
A: Geometric progression
$$\frac{a}{1-x}=a+ax+ ax^2+ax^3+....+..., ~~~|x|<1.$$
Here, $a=1/2$ and $x=1/2$
So $$\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....=\frac{1}{2} \frac{1}{1-1/2}=1=\sum_{n=1}^{\infty}\frac{1}{2^n}$$
