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Let $ m \in \mathbb Z \space with \space m\ge 2,$ $Let\space a_{1},a_{2},a_{3}...\in \left\{0,1,2,3...,m-1\right\}\\Let\space S_{n} = \sum_{k=1}^{n}\frac{a_{k}}{m^{k}},Given \space x\in[0,1],show\space that \space there\space exist \space a_{1},a_{2},a_{3}...\space such\space that \ we\ have \lim_{n\to\infty}S_{n} = x $

I got stuck in this question , I firstly want to pick $a_{k} = x(m-1)$ since this is a geometric series leads to the answer but it does not satisfy the condition that $a_{k}$ must be an interger in [1,m-1]. Anyone can give me some hints for this problem ? Thanks !

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    Don't you wantt $a_i \in \{0,1,2,\ldots,m-1\}$? – 2017-02-13
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    Ooops sorry about that ! – 2017-02-13

2 Answers 2

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You subtract off what you have accounted for already, multiply by another factor of $m$ and round down. So if we want $3.14159$ in base $4$, the integer part is $3$. Then $0.14159 \cdot 4= 0.56636_{10} \lt 1$ so we have $3.0_4$ so far. Then $0.56636_{10} \cdot 4=2.26546_{10}$ so the next place is $2$ and we have $3.02_4$ so far. Then $0.26546 \cdot 4=1$ and a little bit, so we have $3.021_4$ and keep going until we are accurate enough.

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    Thank you very much ! But how can I prove that the limit of sum of this sequence is x ? – 2017-02-13
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    The error at each stage is less than $\frac 1{4^n}$ because you can't round down more than that, so the sum is convergent to $x$ – 2017-02-14
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Let's note $$a_0=0$$ and for every integer $k\geq 0$ $$a_k= {[m^k*x]-m[m^{k-1}*x]} $$ where $[.]$ is the floor function.

then $$S_n = \frac{[m^n*x]}{m^n}$$ wich converge to $x$ when $n\to\infty$ (using $[.]$ definition)