4

Consider the corresponding case for decimal.

  • $9 + 1 = 10$
  • $99 + 1 = 100$
  • $999 + 1 = 1000$
  • ...

We can convince ourselves that $ \overline{9}9 + 1 = 0$ or $-1 = \;...999$

In fact, this can be formalized in the case of floating point arithmetic or p-adic numbers.

Is there a base in which the divergence $1 + 2 + 3 + \dots = - \frac{1}{12}$ makes sense?

To qualify what I mean by base... I tried Factorial base , sequence A007623

  0
  1
 10
 11
 20 
 21 
100 
101 
110 
111 
120 
121 
200 
201

In that case we could imagine the equation $-1_! = \, \dots987654321_!$

cactus314
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  • I don't think $-1=\ldots999$ in floating point. Perhaps you're thinking of two's complement. –  May 03 '14 at 22:13
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    You probably should teconsider being convinced that whatever you mean by $\overline 99$ is equal to -1; even that it has some meaning at all is doubtful. – Mariano Suárez-Álvarez May 03 '14 at 22:34
  • @MarianoSuárez-Alvarez That's why I mention the p-adic numbers where a number like $\dots5$ makes sense. Here $10$ is composite so there may not be 10-adic numbers, if you like, replace it with $7$. – cactus314 May 03 '14 at 22:36
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    @MarianoSuárez-Alvarez $...999999 = -1$ is a perfectly reasonable and correct statement. You can define addition on decimal representations even if they extend infinitely to the left. The result is at the least an abelian group under addition, and if you disallow extending infinitely to the right (after the decimal point) I believe you have a ring. If the base is prime, you get the p-adic number field. – Caleb Stanford May 04 '14 at 05:17
  • In other words, in the ring $\mathbb{Z}_{10}=\varprojlim \mathbb{Z}/10^n \mathbb{Z}$ the serie $\sum_n 9.10^n$ converge to $-1$, and in the ring $\hat{\mathbb{Z}}=\varprojlim \mathbb{Z}/n!\mathbb{Z}$ the serie $\sum_n n.n!$ converge to $-1$. – user10676 May 04 '14 at 22:43
  • @user10676 I am not sure what formally you mean by $\lim\limits_{\leftarrow} \mathbb{Z} / 10^n \mathbb{Z}$. What I mean is if you consider the set of formal infinite decimal expansions $a_0 + 10a_1 + 100a_2 + \cdots$, with $a_i \in {0,1,2, \ldots, 9}$, and you define addition and multiplication (with carrying) in the obvious way, you get a ring. And in that ring, the inverse of $1$ is given by $\ldots99999$. – Caleb Stanford May 04 '14 at 22:47

2 Answers2

1

By definition, the value of an n-digit number $\overline{a_n\ldots a_{_1}a_{_0}}$ in base b is $\displaystyle\sum_{k=0}^na_{_k}b^k$. Now, in order for an infinite series to converge, a necessary albeit by no means sufficient condition is for its general term to tend to $0$, which is equivalent to finding a b for which $\displaystyle\lim_{n\to\infty}~\overline{a_n\ldots a_{_1}a_{_0}}=0$. But this constitutes a problem, since, if $|b|>1$, then the value of the general term diverges towards $\pm\infty$, and if $|b|<1$ then the limit also diverges by oscillation towards $a_{_0}$, whose value passes periodically through any digit from $0$ to $9$. The cases $b=\pm1$ don't fare any better either. Of course, I am fully aware that by “base” you mean something more general than merely “base of numeration”, since you also seem to include p-adic numbers and two's complement, so my answer is partial at best; nevertheless, I think it settles at least one aspect of the biger puzzle, and I'm sure other users will soon contribute with their own perspective. Hope this helps!

Lucian
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1

Unfortunately I see no way to make sense of this if the base is anything normal. In particular, you will notice that in a standard numeral system, like base 2 or factorial base, the value of each digit will always divide the value of the next: $$ \newcommand{\div}{\; \mid \;} 1 \div 10 \div 100 \div 1000 \div 10000 \div \cdots $$ (In base $2$, each power divides the next power of two. In factorial base, each factorial divides the next factorial. This requirement for a numeral system seems necessary, because you need to be able to "carry" when you are adding numbers. Even negative bases have this property.)

The problem with the series $1 + 2 + 3 + \cdots$ is that it includes every natural number. So consider any base, and let $B = 100$, the value of the third digit from the right. Then there are infinitely many natural numbers that are not multiples of $B$. Therefore, in the series, you will infinitely often add a value to the sum which changes the last two digits. Thus, unlike $1 + 2 + 4 + 8 + \cdots$ in base $2$, this sum does not converge digit-wise. That is, it is not the case that each individual digit is eventually constant in the partial sums.

It's possible there is still some way to justify $1 + \tfrac12 + \tfrac13 + \tfrac14 + \cdots = -\tfrac{1}{12}$, but the "base" will have to be very unusual indeed. You could try a fractional base. But once you allow fractional or real bases then you have to decide what it means for a number to be represented in this base, how to make that representation unique, how to add numbers in this base, what digits are possible for each individual place in the representation, and so on. In short it becomes a difficult question. I am still thinking about how to define fractional bases (perhaps someone has a reference?) but I think you should think about your question in this regard and find a way to make it more well-posed.

Caleb Stanford
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