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When we were studying recurring decimals for the first time (that was long ago), I noticed something remarkable whenever I divided a number by 11.
For example $\frac{9}{11}= 0.818181...=0.\overline{81}=0.\overline{(9\times9)} $
We can write it as an infinite sum this way: $$\frac 9 {11}=\sum_{n=1}^{\infty} \left(8 \times 10^{-(2n-1)} +10^{-2n} \right) $$ And then if we try another number between $0$ and $9$, let's choose $7$, we get the same thing:

$\frac 7{11}= 0.636363...=0.\overline{63}=0.\overline{(9\times7)}$

I tried to generalize the previous result and here's what I've got: $$ \frac x{11}= \frac {x-a}{11} + \sum_{n=1}^{\infty} \left(\frac{a-u}{10} \times 10^{-(2n-1)} +u \times 10^{-2n} \right)$$ known that: $ x \equiv a\pmod {11} $ and $ 9a \equiv u \pmod {10}$
I hope it's correct! -crossed fingers-

So, my question is can you help me get a better generalization only using $x$ and $n$ ? And is there any way to prove this result ? If there's any please be as simple as you can and provide explanation. I guess it has something to do with $11$ being the first whole number after $10$ in base $10$.

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    This is a combination of a couple of well-known patterns, which to some of us might be "obvious", but it's put together so thoughtfully that I'm glad to see it already has at least one good answer. – 2017-01-18

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The key is that when $a=1,2,\dots,9$ then $u=10-a$.

It's easier to see it if you write $\frac{1}{11}=0.090909\dots$ and then the question is about multiples of $9$. So $9a=10(a-1)+(10-a)$. And $u\equiv 9a\equiv -a\equiv 10-a\pmod{10}$.

This actually generalizes to other bases.

$$\frac{1}{b+1} = \frac{b-1}{b^2}+\frac{b-1}{b^4}+\cdots$$

And for $a=1,\dots,b$ you get:

$$\frac{a}{b+1} = \frac{(a-1)b+(b-a)}{b^2}+\frac{(a-1)b+(b-a)}{b^4}+\cdots$$