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I came across a problem I would like to ask you about:

Let $x$ be a real number. Iwant to show that $\exists b \in Z$ and integers $ b_1, b_2,b_3, \ldots\in \left\{{0,1,\ldots,9}\right\}$ so that the sequence

$$s_n := b + \sum\limits_{k=1}^n 10^{-k}\cdot b_k$$

converges to the real number $x$.

I can imagine that the sequence looks like ($b=1$ f.ex.) $1,1+\frac{b_1}{10}, 1+\frac{b_1}{10}+\frac{b_2}{100}, \ldots$

But I don't know how to approach this. Is the goal to show that it converges, or that the limit is a real number?

Any help is greatly appreciated!

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Suppose $x$ were the real number $\pi$ which, as you know, has a decimal expansion that starts $3.14159265\dots$. Can you see any relation between this expansion and your question? – Gerry Myerson Oct 4 '12 at 4:03

It’s rather easy to show that any such sequence converges to a real number. Your task is to show that given an arbitrary real number $x$, there is such a sequence that converges to $x$. It’s actually possible to give an algorithm for calculating the integers $b$ and $b_k$ for $k\in\Bbb Z^+$, though of course you will still have to prove that the resulting sequence does converge to $x$. To get you started, I’ll describe the algorithm. (Actually, I’ll describe it for positive $x$; the modifications needed to handle negative $x$ are very straightforward.)

The Algorithm: First, let $b=\lfloor x\rfloor$, the integer part of $x$, and let $x_1=x-b$, so that $0\le x_1<1$. Given $x_k\in[0,1)$ for some $k\in\Bbb Z^+$, let $b_k=\lfloor 10x_k\rfloor$, and let $x_{k+1}=10x_k-b_k$.

To show that this works, show by induction that $s_n\le x$ and that $x-s_n<\dfrac1{10^{n-1}}$ for all $n\in\Bbb Z^+$.

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