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Let's say I take a unit square and cut it up into four equal size squares, each 0.5 side, and say color the upper left red, the lower left blue and the lower right yellow. I can then repeat this same process with the remaining quarter square. After k steps, we have three colored series of squares, each being $\sum_1^n {1\over 4^k}$ and the remaining square is ${1\over 4^n}>0$. This means the sum of all three series will never be more than one and because for any $\delta$ one can easily find such a $n$ that ${1\over 4^n}<\delta$ (logarithm of both sides), it also means it will eventually be more than $1-\delta$ for every delta. Is this an actual proof for $\sum {1\over 4^k} = {1\over 3}$ or is this wrong somewhere?

person
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chx
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1 Answers1

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You showed $3 \sum_{k=1}^n 4^{-k}=1-4^{-n}$, which is after a small rearrangement a proof of $\sum_{k=1}^\infty 4^{-k}=1/3$.

Ian
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  • What is the "small rearrangement"? You can certainly get that last equation by taking limits as $n\to\infty...$ – Adam Rubinson Apr 29 '22 at 21:25
  • @AdamRubinson I believe the "small rearrangement" is the shifting of the $3$ to the other side after the limit as $n \to \infty$ has been computed. –  Apr 29 '22 at 21:35
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    That's correct. If you're being really pedantic, the proof isn't done until you actually send $n \to \infty$ and move the 3 over. – Ian Apr 29 '22 at 21:47