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(I've never posted on here before, so apologies for any formatting problems)

I had always noticed this property of the decimal form of $1/7$ ( $0.14285714...$ ) where the decimals went $14$ , then $28$ , then $57$ , which is almost exactly the formula $f(x)=7*2^x$ , starting with $f(1)$ - except for $57$ , which is $1$ more than $56$. However, if you add the next number in the series, $f(4)=112$ , to the tail end of $56$, then you get $57$ and are left with the second $1$ and the $2$ to add to the end of the decimal, which is exactly where the $71$ lands in the $.14285714...$ string of digits. If you keep doing this with the next number in the formula, $f(5)=224$ , and add to the end of $112$ , you get $2+2$ , which accounts the $4$ and the $2$ , and so on.

I continued this pattern and found that this process of adding digits together works all the way up to the $45th$ digit , which is where I stopped. Is this simply a coincidence or is there a reason for this to occur? And if there really is something behind this (which I'm sure there is, does anyone know why this happens? I would appreciate any attempt at solving this.

Work for said question (apologies if it's messy, I did this work in a cramped notebook lol) I know this seems very confusing, but if anyone needs me to clarify, I will do my best.

Bill Dubuque
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dckjavis
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1 Answers1

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It is Geometric Progression with the Co-Incidence that $50 = 7 \times 7+1$

When we are appending 2 Digits , we are using $100^n$ Denominator , while the Numerator is what you gave $7 \times 2^n$

Adding up the terms for all $n$ will give this :
$X = 7 \times \sum_{n=1}^{\infty}x^n = 7 \times x/(1-x)$ , where $x = 2/100=1/50$

We can then see that $x/(1-x)=1/49$
Hence $X=7 \times 1/49 = 1/7$

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