= Solution
For $q\in\mathbb Q\setminus\{0\}$ write $q=p^k u/v$, where $k\in\mathbb Z$ and neither integer $u$ nor integer $v$ is divisible by $p$. Define the <P-adic valuation> $v_p(q)=k$ and the <p-adic absolute value> $|q|_p=p^{-k}$, with $|0|_p=0$. \b[The p-adic metric] is
$$
\boxed{d_p(a,b)=|a-b|_p.}
$$
Factoring out the smaller power of $p$ from two rational numbers shows $v_p(q+r)\geq\min\{v_p(q),v_p(r)\}$, with the usual convention $v_p(0)=+\infty$. Consequently the <ultrametric inequality> gives
$$
d_p(a,b)=|(a-c)+(c-b)|_p\leq\max\{d_p(a,c),d_p(c,b)\}.
$$
Together with symmetry, positivity and separation of points, this verifies that $d_p$ is a <metric>.
For the given geometric partial sums, use the finite <geometric series> identity. Since $p$ does not divide $p-1$,
$$
a_n=\frac{p^n-1}{p-1},\qquad \left|a_n+\frac1{p-1}\right|_p=\left|\frac{p^n}{p-1}\right|_p=p^{-n}\longrightarrow0.
$$
Thus \b[the limit is the rational number] $\boxed{1/(1-p)}$, even though the partial sums diverge in the ordinary real <metric>.
If $|a|_p>|b|_p$, the <ultrametric inequality> gives $|a+b|_p\leq|a|_p$. Applying it to $a=(a+b)-b$ gives $|a|_p\leq\max\{|a+b|_p,|b|_p\}$. Since $|b|_p<|a|_p$, this forces $|a+b|_p=|a|_p$. Interchanging $a,b$ handles the other unequal case, proving
$$
\boxed{|a+b|_p=\max\{|a|_p,|b|_p\}\quad\text{when }|a|_p\ne|b|_p.}
$$
Finally, let $b$ be outside the <open ball> $B(a,\delta)$, so $d_p(a,b)\geq\delta$. If $d_p(b,c)<\delta$, the preceding unequal-norm identity gives $d_p(a,c)=d_p(a,b)\geq\delta$. Therefore $B(b,\delta)$ also lies outside $B(a,\delta)$. Its complement is <open>, so \b[every p-adic open ball is also closed], that is, a <clopen set>.
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