Geometric series convergence in the p-adic numbers

Question

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.

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).