3

If we have $A = \displaystyle \sum_{j=0}^{\infty} a_j$ with $a_j \in \mathbb{Q}$ and $A$ converges both in the reals, and in some p-adic $\mathbb{Q}_p$, is there any connection between the sum being rational in $R$ vs. $\mathbb{Q}_p$?

Thanks!

Niklas
  • 341
  • 1
    Related and food for thought: https://math.stackexchange.com/q/3264177/96384, https://math.stackexchange.com/q/3059249/96384, https://math.stackexchange.com/q/2300172/96384. Note in particular the example $$\displaystyle (1+(-\frac78))^\frac13 = \sum_{k=0}^\infty \binom{1/3}{k}(-7/8)^k $$ which as a series in $\mathbb R$ converges to $0.5 \in \mathbb Q$, but as a series in $\mathbb Q_7$ converges to a $7$-adic number $\notin \mathbb Q$. – Torsten Schoeneberg Mar 22 '20 at 04:22

2 Answers2

5

No, that's the point of different norms, there is a sequence $(a_n)$ converging to $1$ in one and to $0$ in the other, thus letting $z_n=a_{f(n)} x_n+ (1-a_{f(n)})y_n$ where $f(n)\to \infty$ fast enough you get that $\lim z_n$ is $\lim x_n$ in 1st norm and $\lim y_n$ in 2nd norm.

Concretely $$a_n = \frac{1}{1+p^n}$$

reuns
  • 79,880
  • 1
    But I think OP was asking about infinite series, and specifically for an example of a series convergent in both norms that has rational sum in one and irrational sum in the other. I hope they see that your example can be converted to something of the sort they desire. – Lubin Mar 12 '20 at 19:36
  • Thank you for the answer! How would this conversion that Lubin mentioned be made? And would not the p-adic sum diverge as the term $1-a_{f(n)} = \frac{p^n}{1+p^n}$ which approaches 0, while $\frac{1}{1+p^n}$ would produce an infinite sum of terms with p-adic order 0? Thank you. – Niklas Mar 13 '20 at 09:09
  • 2
    Well, any time you have a convergent sequence ${a_n}$, you can get from it a convergent series $\sum_nb_n$ by taking $b_n=a_n-a_{n-1}$ (and making special arrangements for the first $a_i$), so the problem is to find a sequence ${a_n}$ with a (say) rational real limit and an irrational $2$-adic limit. – Lubin Mar 18 '20 at 21:15
2

Thinking over the answer of @reuns of several days ago, I decided that perhaps a more complete answer would be called for. I’ll take all my examples from two different metrics on $\Bbb Q$: the archimedean, which gives the completion $\Bbb R$, and the nonarchimedean $5$-adic, which gives $\Bbb Q_5$.

First, let’s look at a series with the same, rational limit in the two metrics, such as$$\sum_{n=0}^\infty\left(\frac57\right)^n\,.$$ Here, the common ratio is small in both the real and the $5$-adic metrics, so the high-school formula applies, $a/(1-r)$, to give a limit of $7/2$ in both cases. It becomes clear just from this example that a geometric series of rationals that’s convergent in any two metrics will have the same, rational limit there. So we need a more outré example to satisfy your demand of a series that has a rational limit in one metric, irrational in the other.

Thus we need to start with perhaps our favorite irrational $p$-adic number, like $\sqrt{-7}\in\Bbb Q_2$ or $\sqrt7\in\Bbb Q_3$. But I promised you a five-adic example: I choose a square root of $-1$, of which there are of course two: one that’s $\equiv2\pmod5$, the other $\equiv3\pmod5$.

I don’t know whether you’re familiar with the semistandard notation for $p$-adic numbers, using a semicolon instead of a decimal point: in this notation, we get $$ i=\dots40423140223032431212; $$ to twenty places, and you read it right to left. It means $$ i=2+5+2\cdot5^2+5^3+3\cdot5^4+4\cdot5^5+\dots $$ Negative powers of $5$ get put to the right of the semicolon, as you’d expect, so that $19/5$ is denoted $4;3$. If we were considering two different prime radices, I’d put the prime as a subscript to the semicolon: $4;_53\>$.

As I suggested in my second comment, you can go back and forth between a convergent sequence and the corresponding convergent series. If you try that with the $5$-adically convergent series for $i$ above, you’ll get a sequence (the partial sums) that is unbounded in the real sense, not what we’re looking for at all. So I’ll use a trick to get a $5$-adically convergent series whose partial sums approach $0$ in the real sense. For this, I’ll look at the expansion of $-i=1/i$. This is: $$ -i=\dots04021304221412013233; $$ You look at this as a sum, in which the first partial sum $s_0=3$, the second is $s_1=3+3\cdot5$, the third is $s_2=3+3\cdot5+2\cdot5^2$, etc. Then this sequence is $5$-adically convergent to $-i$, and unbounded in the real sense. Consequently, the sequence $\{1/s_n\}$ is convergent to zero in the real sense, and to $i$ in the $5$-adic sense.

Now it’s just a matter of taking the associated sum $\sum_{n=0}^\infty a_n$ where $a_0=1/s_0=1/3$, and for $n>0$, $a_n=1/s_n\,-\,1/s_{n-1}$. And there you are.

Lubin
  • 65,209