Geometric series convergence in the p-adic numbers

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.

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).
• 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. May 28, 2017 at 15:15
• 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 $p|a$). May 28, 2017 at 15:20
• This is a really good example of the one series giving two different limits, depending on the $p$. Plus 1. May 29, 2017 at 21:35