Prove that 1/2 + 1/4 + 1/8 ....... = 1 I've often heard that instead of adding up to a little less than one, 1/2 + 1/4 + 1/8... = 1. Is there any way to prove this using equations without using Sigma, or is it just an accepted fact? I need it without Sigma so I can explain it to my little sister.
It is not a duplicate because this one does not use Sigma, and the one marked as duplicate does. I want it to use variables and equations.

$A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\cdots+\dfrac{1}{2^{n+1}}$
$A=\dfrac{2^n}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\dfrac{2^{n-2}}{2^{n+1}}+\cdots+\dfrac{1}{2^{n+1}}$
$2A=\dfrac{2^{n+1}}{2^{n+1}}+\dfrac{2^{n}}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\cdots+\dfrac{2}{2^{n+1}}$
Then
$A=2A-A$
$=\left(\dfrac{2^{n+1}}{2^{n+1}}+\dfrac{2^{n}}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\cdots+\dfrac{2}{2^{n+1}}\right)-\left(\dfrac{2^n}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\dfrac{2^{n-2}}{2^{n+1}}+\cdots+\dfrac{1}{2^{n+1}}\right)$
$=\dfrac{2^{n+1}}{2^{n+1}}-\dfrac{1}{2^{n+1}}=1-\dfrac{1}{2^{n+1}}$
If the expression $A$ has more fractions, $n+1$ will be larger, which means $\dfrac{1}{2^{n+1}}$ should be closer to zero, which means $A$ converges to $1$.
