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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.

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    $\begingroup$ People say all sorts of strange things. A fact is that $\sum_{n=1}^\infty 2^{-n}$ converges to $1$. $\endgroup$ – Angina Seng May 12 '18 at 14:07
  • $\begingroup$ The concept that makes sense out of such expressions is called infinite series and convergence of such series. $\endgroup$ – zipirovich May 12 '18 at 14:12
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    $\begingroup$ I don't agree this question is a duplicate: the O.P. asks explicitly an explanation that a child can understand. I don't think series are a valid answer. $\endgroup$ – Bernard May 12 '18 at 14:24
  • $\begingroup$ @Bernard The sentence about explaining it to his/her little sister was added by the OP after the question had been closed. $\endgroup$ – Angina Seng May 12 '18 at 14:27
  • $\begingroup$ Well, I voted to reopen it, having seen this addition. $\endgroup$ – Bernard May 12 '18 at 14:27
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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 $n \to \infty$.

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

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

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

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 $1$.

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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.

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  • $\begingroup$ Not always. Convergence has to be justifies first. $\endgroup$ – zipirovich May 12 '18 at 14:13
  • $\begingroup$ You are correct, but in this case convergence is well defined.. I shall however add your point. Thank you for telling me :) $\endgroup$ – The Integrator May 12 '18 at 14:14
  • $\begingroup$ You're correct now, but you didn't say that in your original post. What you did say makes it sound like this is true for all geometric series. Since the OP clearly is not familiar with the concept, we shouldn't post "answers" that provide wrong and/or misleading information to those who are trying to learn something new to them. $\endgroup$ – zipirovich May 12 '18 at 14:16
  • $\begingroup$ @zipirovich Thank you. I have rectified my error, would you please check it. Feel free to edit it if you find it incorrect. $\endgroup$ – The Integrator May 12 '18 at 14:17
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    $\begingroup$ As per your invitation, I took the liberty to add one more comment. Great teamwork! :-) $\endgroup$ – zipirovich May 12 '18 at 14:22

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