Prove that 1/2 + 1/4 + 1/8 ....... = 1

Question

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.

Answer 1

For physical intuition, so you can explain it to your little sister, I will use a 1m long ruler.

Take the ruler an divide it into two equal parts:

$$1=\frac{1}{2}+\frac{1}{2}$$

Take one of the parts you now have, and again divide it in half.

$$=\frac{1}{2}+\frac{1}{4}+\frac{1}{4}$$

Take one of the smaller parts you now have, and again divide it in half.

$$=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}$$

Repeat. In general for $n$ a positive integer,

$$=\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n} \right)+\frac{1}{2^n}=1$$

So,

$$\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}=1-\frac{1}{2^n}$$

As we let $n$ become a really big (positive) integer, note the sum gets closer and closer to $1$, because $\frac{1}{2^n}$ gets really close to zero (the smallest part of the ruler you have left over gets close to 0 meters in length). We say the sum converges to $1$ in the limit that $n \to \infty$.

Answer 2

It is an infinite Geometric series .

The sum in an infinite geometric series is given by $S = \frac{a_1}{1-r}$ where $a_1$ is the first term and $r$ is the common ratio.

In your case ;

$\frac12+\frac 14+\frac 18 \cdots = \sum_{n=1}^{n=\infty}\big(\frac12\big)^n = \frac{\frac12}{1-\frac12}= \frac{\frac12}{\frac12} = 1$

EDIT 1:

As noted down in the comments, convergence is not always guaranteed by the above formula is mentioned for that i recommend you check out Convergence tests and infinite series

EDIT 2:

In particular, for geometric series of the form $$\sum_{n=0}^{\infty} ar^n=a+ar+ar^2+ar^3+\cdots+ar^n+\cdots$$ such a series converges, meaning it has a well-defined sum, if and only if $|r|<1$. (And you shouldn't take this for granted — there's an actual proof for this fact.) In this example, $r=\frac12$ and $|r|=\frac12<1$, so it does converge.

Answer 3

$A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\cdots+\dfrac{1}{2^{n+1}}$

$A=\dfrac{2^n}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\dfrac{2^{n-2}}{2^{n+1}}+\cdots+\dfrac{1}{2^{n+1}}$

$2A=\dfrac{2^{n+1}}{2^{n+1}}+\dfrac{2^{n}}{2^{n+1}}+\dfrac{2^{n-1}}{2^{n+1}}+\cdots+\dfrac{2}{2^{n+1}}$

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 $1$.