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I am currently working on the completeness of metric spaces, so I studied the following theorem:

If $E$ is a Banach space then any absolutely convergent series is convergent.

Since $\mathbb Q$ is not complete, I wanted to illustrate this theorem with an absolutely convergent series on $\mathbb Q$ which wouldn't converges:

$$\sum_{n=0}^\infty \vert r_n\vert\in\mathbb Q\quad\text{ and }\quad \sum_{n=0}^\infty r_n\in\mathbb R\setminus\mathbb Q \quad (\text{ with $r_n\in \mathbb Q$}).$$

Is was not able to find such series. Is there an example like it (as simple as possible ?).

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  • $\begingroup$ To exclude trivial cases i think that $r_i\in\Bbb Q$ for all $i$ must be required. $\endgroup$ – Martín-Blas Pérez Pinilla Oct 3 '16 at 9:23
  • $\begingroup$ $r_n = 1/n! {}{}{}$. And that any Cauchy sequence in a Banach space converges is a definition, not a theorem. A point $p$ in a normed space is defined by its neighborhoods (the balls $B_p(\epsilon) = \{x \in B, \|x-p\| < \epsilon\}$) and a Cauchy sequence has a "limit" in the sense that it gets closer and closer to it, but the truth is that it is the opposite : the limit is defined precisely by the sequence of elements getting closer and closer to it $\endgroup$ – reuns Oct 3 '16 at 9:45
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    $\begingroup$ @user1952009 $\sum 1/n!=e$ so it doesn't not absolutely converges on $\mathbb Q$ right ? $\endgroup$ – E. Joseph Oct 3 '16 at 9:46
  • $\begingroup$ @E.Joseph ok, your question is so weird. In a normed/Banach space, nobody cares it the sequence of norms converges in $\mathbb{Q}$ or not, so this is special to $\mathbb{Q}$ (and has nothing to do with Banach spaces) $\endgroup$ – reuns Oct 3 '16 at 9:51
  • $\begingroup$ @user1952009: The question is not special to $\mathbb{Q}$. It can be asked in any ordered field, and the answer is that absolute convergence implies convergence is equivalent to sequential completeness. It can also be asked in any ($\mathbb{R}$)-normed abelian group, and the property is equivalent to completeness of the associated metric. See e.g. math.stackexchange.com/questions/111164/…. $\endgroup$ – Pete L. Clark Oct 3 '16 at 15:31
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We have that $$\sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+1)}= \sum_{n=1}^{\infty}\left(\frac{(-1)^n}{n}+\frac{(-1)^{n+1}}{n+1}\right)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n+1}\\ =-\ln 2+(-\ln 2+1)=1-2\ln 2\in \mathbb R\setminus\mathbb Q $$ and $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)}= \lim_{N\to+\infty}\sum_{n=1}^N\left(\frac{1}{n}-\frac{1}{n+1}\right)=\lim_{N\to+\infty}\left(1-\frac{1}{N+1}\right)=1\in \mathbb{Q}.$$

P.S. See here for $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=-\ln 2$.

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  • $\begingroup$ That is precisely what I was looking for, thanks ! Do you konw where can find a proof of this statement ? $\endgroup$ – E. Joseph Oct 3 '16 at 9:48
  • $\begingroup$ @E. Joseph I edited my answer with the details. $\endgroup$ – Robert Z Oct 3 '16 at 9:53
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A trivial example can be constructed by taking any $\sum_{i=2}^{\infty}r_i$ convergent and absolutely convergent on $\mathbb{Q}$ and add $r_0=5+(\pi-5)/2$, $r_1=(\pi-5)/2$. You get $$|r_0|+|r_1|=5+(\pi-5)/2-(\pi-5)/2=5$$ but you get $$r_0+r_1=5+(\pi-5)/2+(\pi-5)/2=5+\pi-5=\pi$$.

Note: You don't actually point out you want the original series made up of elements of $\mathbb{Q}$ but I realized that must be your intention.

But we can easily fix this.

Take $\{a_i\}_{i=0}^\infty$ such that $\sum_{i=0}^\infty a_i=r_0$ and $a_i>0$ for all $i$. Then take $\{b_i\}_{i=0}^\infty$ such that $\sum_{i=0}^\infty b_i=r_1$ and $b_i<0$ for all $i$. Then put alternate them ($c_{2n}=a_n$ and $c_{2n+1}=b_n$).

Notice that since everything converges absolutely you get the right answer.

Consider $$S_{2n}=\sum_{i=0}^{2n-1}c_n=\sum_{i=0}^{n-1} a_i+\sum_{i=0}^{n-1}b_i$$ And so since both converge absolutely you get the correct sum without absolute values and since you get $|a+b|=-(|a|+|b|)$ when both $a,b<0$. You get what you would expect in the absolute case aswell.

Further edit: A visual interpretation is that $r_0$ is a mid point between a larger rational and smaller irrational and the $r_1$ is a fix upwards to the rational if it's positive or downwards to the irrational if it's negative.

Yet another edit: You can actually show that all series meeting the requirements of the OP (with the restriction of $r_i$ to $\mathbb{Q}$ must be of this type.

First note that if $\sum_{i=0}^{\infty}|r_i|$ is in $\mathbb{Q}$ and $\sum_{i=0}^\infty r_i$ is not, then there must be infinitely many positive and negative $r_i$'s.

This follows easilly in two steps: First all can't be positive (otherwise both sums are the same) or negative (otherwise one sum is just the opposite of the other so both are in or out of $\mathbb{Q}$). Second if you have only finitely many positive/negative remove them, (their sum is in $\mathbb{Q}$) and the rest follows by first part.

Now we know that there are infinitely many positive and negative $r_i$'s and the whole sum is absolutely convergent. Thus all subsums are absolutely convergent. Now just split the sum of the positive terms and negative terms (those both converge by absolute convergence) and give you your $r_0$ and $r_1$ which have the property that $r_1<0$, $r_0>0$.

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    $\begingroup$ I think that $r_i\in\Bbb Q$ for all $i$ must be required. $\endgroup$ – Martín-Blas Pérez Pinilla Oct 3 '16 at 9:22
  • $\begingroup$ @Martín-BlasPérezPinilla Yeah I realized it myself once I posted. I think this works properly but would be happy if someone double checks. The fact that everything converges absolutely makes it work correctly I belive. $\endgroup$ – DRF Oct 3 '16 at 9:29
  • $\begingroup$ @Martín-BlasPérezPinilla Essentially the idea is take something in $\mathbb{Q}$ (5) and something not in $\mathbb{Q}$ ($\pi$) look at the middle point and add upwards from it by absolute value and downwards from it without absolute value. $\endgroup$ – DRF Oct 3 '16 at 9:31
  • $\begingroup$ @DRF Since $\sum a_i\in \mathbb Q$ and $\sum b_i\in \mathbb Q$, we have $\lim S_{2n}\in \mathbb Q$ as well right ? So I don't get why $S$ would not be convergent in $\mathbb Q$... $\endgroup$ – E. Joseph Oct 3 '16 at 9:42
  • $\begingroup$ @E.Joseph the sum of neither $a_i$ or $b_i$ is in $\mathbb{Q}$ actually. Neither $r_0$ or $r_1$ is. $\endgroup$ – DRF Oct 3 '16 at 9:50
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Let be $$\sqrt 2 = 1.0110101\dots = \sum_{i=0}^\infty a_i2^{-i}$$ with $a_i = 0,1$ and take $$\Sigma = 1 - \frac12+\frac14+\frac18-\frac1{16}+\cdots$$ (negative sign if the corresponding binary digit is $0$)

Then, as $$2 = 1.1111111\dots = 1.0110101\dots + 0.1001010\dots = \sqrt 2 + (2-\sqrt 2),$$ we have $\Sigma = \sqrt 2 - (2-\sqrt 2) = 2\sqrt 2 -2$ but $$1 + \frac12+\frac14+\frac18+\frac1{16}+\cdots = 2.$$

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The relationship between convergence and absolute convergence in an ordered field is studied in this note, which has its genesis in a question that was asked on this site (by me; it was answered by my coauthor, N.J. Diepeveen).

The case of subfields of $\mathbb{R}$ makes for a nice undergraduate exercise (as seems to be happening here!): see Theorem 1. Or let me explain: let $x \in [-1,1] \setminus \mathbb{Q}$. Then there is a sequence (in fact a unique sequence, since $x \notin \mathbb{Q}$, but this is not needed for what follows) $\epsilon_n \in \{ \pm 1\}$ such that

$x = \sum_{n=1}^{\infty} \frac{\epsilon_n}{2^n}$.

The way you get $\epsilon_n$ is by a simple inductive process: for $n \geq 0$, having chosen $\epsilon_1,\ldots,\epsilon_n$ already, we choose $\epsilon_{n+1}$ to be $1$ if $\sum_{k=1}^n \frac{\epsilon_k}{2^k} < x$ and $-1$ if $\sum_{k=1}^n \frac{\epsilon_k}{2^k} > x$. A little thought shows that this converges to $x$. Now notice that no matter what the sign sequence $\epsilon_n$ is, the absolute series is

$\sum_{n=1}^{\infty} \frac{1}{2^n} = 1 \in \mathbb{Q}$.

Note that this argument works equally well with $\mathbb{Q}$ replaced by any proper subfield $F \subsetneq \mathbb{R}$, or in fancier terms, any incomplete Archimedean ordered field.

Note also that this is essentially the same answer as @Martín-Blas Pérez Pinilla, but I wanted to give my take on it.

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