# Relation between rationality of same infinite series convergent both in $\mathbb{R}$ and $\mathbb{Q}_p$

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!

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

• 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. Mar 12, 2020 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. Mar 13, 2020 at 9:09
• 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. Mar 18, 2020 at 21:15

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.

• Fantastic, thank you! Mar 19, 2020 at 23:21