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Expand the expression $1-\frac{1}{2}(1-\frac{1}{2}(1-\frac{1}{2}))\\ =1-\frac{1}{2}+(\frac{1}{2})^2(1-\frac{1}{2})\\ =1-\frac{1}{2}+(\frac{1}{2})^2-(\frac{1}{2})^3\\$

The expression represents the partial sum $\sum_{n=0}^{3}(-1)^n\cdot(\frac{1}{2})^n\\$

That means the following expression is an infinite alternating geometric series representing $\sum_{n=0}^{\infty}(-1)^n\cdot(\frac{1}{2})^n$

$1-\frac{1}{2}(1-\frac{1}{2}(1-...)))$

Let $x$ represent the sum of the infinite series

$ x =1-\frac{1}{2}(1-\frac{1}{2}(1-...)))$

Then substitute $x $ in for the expression $ 1-\frac{1}{2}(1-...)$ above.

It follows

$x=1-\frac{1}{2}x$

Solving for $x $

$x=\frac{2}{3} $

It is known that by the following $\sum_{n=0}^{\infty}(-1)^n\cdot{r}^n=\frac{1}{1+r}$

From this known fact it also follows $\sum_{n=0}^{\infty}(-1)^n\cdot(\frac{1}{2})^n=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}$

See Formula for Alternating Geometric Series.

Ramon Munoz
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    It works, but it really isn't "new" - this seems to me to really be equivalent to the usual argument. Also note you have only shown that if the series converges then it converges to 1/3. But you would need to first show that the series does converge in order for your manipulations to be valid. – Nate Eldredge Oct 04 '19 at 02:44
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    I have been using this formula from school. – Bijay Oct 04 '19 at 02:44
  • @BJKShah The formula at the bottom is what I use in school, too. The representation of the geometric series and the substitution for x is all too new. – Ramon Munoz Oct 04 '19 at 03:00
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    @NateEldredge I have depended on previous work for the example of a converging series, I don't know how I could otherwise prove the series converging. – Ramon Munoz Oct 04 '19 at 03:04
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    @NateEldredge Would I have to offer some sort of convergence test based on the characteristics of the original expression? – Ramon Munoz Oct 04 '19 at 03:19
  • @NateEldredge I understand now what you are saying. The example I provided does not converge because |r|>1. I can edit my explanation to include a value of |r|<1. – Ramon Munoz Oct 04 '19 at 04:28
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    Standard method: $S=1-\frac12+\frac14-\frac18+\cdots=1-\frac12(1-\frac12+\frac14-\cdots)=1-\frac12S \Rightarrow S=\frac23$. See also Alternating series test – farruhota Oct 04 '19 at 04:37
  • @farruhota I had no idea that's what it's called. In https://en.wikipedia.org/wiki/Geometric_series Subtracting two self-similar series seems to be the method commonly used while obtaining $S-rS=1$ as opposed to what is obtained by the manipulation above: $1+rS=S$. The self-similarity principle is illustrated in the standard method you and I provide, meanwhile the standard method does not feel artificially contrived. – Ramon Munoz Oct 04 '19 at 05:18
  • The idea you're used is a fixed point argument, i.e. you're looking for $y=\lim_{n\to\infty} f^n(x)$, where $f^n$ shall be n-times the composition of $f$ with itself. So you have $y=y \Leftrightarrow y=f(y)$. Though it is interesting to see it in a sum-setting – Sudix Oct 04 '19 at 07:11

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