Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
Hint: If you have defined $a_1,a_2,\dots,a_n$, let $a_{n+1}=1$ if $\frac{a_1}{2}+\cdots+\frac{a_n}{2^n}+\frac{1}{2^{n+1}}\le x$, and $0$ otherwise. –  André Nicolas Nov 12 '12 at 7:01
This is clearly the way to go, via an inductive definition of the $a_k$. Should also throw the bounds into the inductive definition of sum-so-far < x < sum-so-far +1/2^n or the like. +1 for comment. –  coffeemath Nov 12 '12 at 7:16
Thank you, so I use induction? Is this similar to Cantor's diagonal argument? –  Alti Nov 12 '12 at 20:46
It's not like the diagonal argument. What Andre's remark is about is how to construct the sequence of $a_n$ from the number $x$. If you do some cases, you'll see that (given $a_1,...,a_n$ are correct), the definition gives the correct next base 2 digit $a_{n+1}$. –  coffeemath Nov 12 '12 at 23:53
Can you give me an example? I'm sorry, I feel it's not clicking yet. –  Alti Nov 13 '12 at 1:08
show 1 more comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
So I would let $x$ be in $[0,1]$, and let $(a_n)$ be a sequence such that $a_n$ is in {0,1} for all n. I'm not sure what you mean by constructing the sequence $a_n$ from $x$, though. $x$ is a number, so I would construct a number of zeroes and ones? –  Alti Nov 12 '12 at 6:53
Yes, each particular $x$ would have its own particular sequence of 0's and 1's, which are obtained from the base two decimal version of $x$. For example I think that 1/3 has the base two expansion $0.01010101...$ where the digits repeat in blocls of two. –  coffeemath Nov 12 '12 at 7:11
Thank you for all your help! –  Alti Nov 14 '12 at 3:49
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 $x$. –  coffeemath Nov 14 '12 at 4:08
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.