I am not satisfied with my understanding of this type of question:

let $a \in \mathbb Q$, for which primes p does the series $\sum_{k = 0}^{\infty} a^k $ converge in the p-adic numbers $\mathbb Z_p$ ?

Find the sum of the series in each case.

Let's just say $a = 15/7$, then I understand that the sum converges for $p = 3$ and $p = 5$ since $|a|_3 = 1/3 < 1$ and $|a|_5 = 1/5 < 1$

But when the sum converges isn't it always just $\sum_{k=0}^{\infty} \frac{15}{7} = \frac{1}{1-\frac{15}{7}} = -\frac{7}{8}$ ? I would like someone to confirm or deny this assertion.


1 Answer 1


As you noted the sum of the geometric series $\sum_{k=0}^{\infty}r^n = \frac{1}{1-r}$ whenever the series converges. In this case we get the same rational number answer whether it's convergence over the reals or over the $p$-adics. But this is not always the case even when the answer is a rational number. An easy example can be found by considering $$ (1+x)^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{-1/2}{k} x^k $$

Take $x=-\frac{24}{49}$ so that $(1+x)=(\frac{25}{49})$. Over $\Bbb{R}$ we calculate $(\frac{25}{49})^{-\frac{1}{2}}= \frac{7}{5}$, but over $\Bbb{Q}_2$ the series gives the root $(\frac{25}{49})^{-\frac{1}{2}}= -\frac{7}{5}$, because the root calculated from the series in $\Bbb{Q}_2$ must be $\equiv 1\bmod{4}$ (like knowing that the root in $\Bbb{R}$ must be positive).

  • $\begingroup$ Thanks, I assume there are examples where a series converges to different fractions in, let's say, $\mathbb Q_2$ and $\mathbb Q_3$ (?). But then why does the question (in this case the original question was $a = 2016$) ask for the limit in each case, when the limits are the same? Maybe it's just testing us to see if we understand that the limit is the same. $\endgroup$ May 28, 2017 at 15:15
  • $\begingroup$ I think one of the things they are trying to get at is making sure you understand that the geometric series only converges for $|a|_p<1$ and what that entails (basically $p|a$). $\endgroup$
    – sharding4
    May 28, 2017 at 15:20
  • $\begingroup$ This is a really good example of the one series giving two different limits, depending on the $p$. Plus 1. $\endgroup$
    – Lubin
    May 29, 2017 at 21:35

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