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I completed the first part of this question, which is stated [here] Series convergence for sequence of zeroes and ones

I showed that the sequence of partial sums $S_n$ is bounded $0\le S_n\le1$, so as n goes to infinity, the sequence of partial sums converges to some value between 0 and 1. This is the second part of the question: Let $x\in[0,1]$. Prove that there exists a sequence $(a_n)_{n\in N}$ such that $x=\sum_{n=1}^\infty \frac{a_n}{2^n}$ This is the binary expansion of $x$. Isn't this part of the question basically stating what I proved in the first, i.e. I showed that the sequence of partial sums is between 0 and 1. Or am I misunderstanding the question?

Any clarification is appreciated.

Thank you.

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    I put some details in the answer I gave.2012-11-13

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In this part you are given x and have to show the existence of the sequenced $a_n$.

So this means, for each fixed $x$ in $[0,1]$, you have to do two things:

1) show how to construct the sequence $a_n$ from the number $x$, and

2) show that the resulting sum $\sum a_n/2^n$ actually converges to $x$.

In the previous part of the question you showed such sums converge, but in this latter part of the question you are to show for each $x$ you can find an appropriate sequence which converges with its sum being that particular $x$.

EDIT: The OP Alti has asked for details on the construction of the $a_n$ from the number $x \in [0,1]$. First note that the particular number $x=1$ has the expression in which all $a_n=1$, i.e. $1=1/2+1/4+1/8+...$, so that we may assume in fact that $X \in [0,1)$, the half-open interval where $0 \le x < 1.$

To get the construction started, we use that $[0,1)=[0,1/2) \cup [1/2,1)$ where the union is disjoint. We let $a_1=0$ if $x \in [0,1/2)$ and $a_1=1$ if $x \in [1/2,1)$. Note for this "base case" of the construction that we have $a_1/2 \le x < a_1/2+1/2$, which may be restated as $x \in [a_1/2,a_1/2+1/2)$ To construct the next $a$, which is $a_2$, we use that $[a_1/2,a_1/2+1/2)=[a_1/2,a_1/2+1/4) \cup [a_1/2+1/4,a_1/2+1/2),$ the union again being disjoint. We then define $a_2=0$ if $x$ is in the first half of the above disjoint union, and $a_1=1$ if x lies in the second half of the above union.

For notation of left and right endpoints, let v_n=a_1/2+a_2/4+...+a_n/2^n$, so that $v_n$ is the $n^{th}$ partial sum of the series we are constructing. Then provided we have inductively constructed each of $a_1,a_2,...,a_n$ we have at that stage that $x \in [v_n,v_n+1/2^n).$ For constructing $a_{n+1}$ we use the disjoint union $[v_n,v_n+1/2^n)=[v_n,v_n+1/2^{n+1}) \cup [v_n+1/2^{n+1},v_n+1/2^n).$ This is another disjoint union, and we let $a_{n+1}=0$ if $x$ lies in the left half and $a_{n+1}=1$ if $x lies in the right half.

Convergence of the partial sums to x$ can be based on the nested interval theorem, or on using the partial sums of the series and also the lengths of the constructed half-open intervals in the proof.

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    Alti: An interesting topic to look up is "bisection method" or some title like that. Basically one takes an interval and keeps cutting it in half, and focussing on where your point $x$ is, then cutting that part in half, and so on. For $x \in [0,1]$ you can keep track of the bisections and get base 2 digits of $x$.2012-11-14