For which algebraic numbers $\alpha$ is there a valuation on the number field ${\mathbb {Q}}(\alpha)$ for which the infinite series $\sum\_{n=0}^\infty \alpha^n$ converges to $1/(1-\alpha)$?



Geometric series in algebraic number fields
This holds precisely for elements which aren't roots of unity. Indeed, by the product formula we have $\prod\_v|\alpha|\_v=1$, where the product runs over all (finite and infinite) primes. This shows that either all $|\alpha|\_v$ are equal to $1$, or at least one of them is smaller than $1$. If all $|\alpha|\_v$ are equal to $1$, then $\alpha^n$ doesn't tend to zero in any valuation, so $\sum\alpha^n$ cannot converge. This happens precisely when $\alpha$ is an algebraic integer (that property follows from having $|\alpha|\_v\leq 1$ for all the finite places $v$) with all conjugates of absolute value $\leq 1$ (infinite places), which by a [theorem of Kronecker](https://mathoverflow.net/q/10911/30186) characterizes roots of unity.


If $|\alpha|\_v<1$ for some place $v$, then $\sum\alpha^n=\frac{1}{1-\alpha}$ in the valuation induced by $v$, regardless of whether it is a finite or infinite place.


The above answer is assuming you consider valuations/absolute values coming from all places in your question. If you only consider finite places, then there isn't such a complete characterization. Rather, you just want $\alpha$ to have positive valuation at some finite place, which you can characterize by saying $1/\alpha$ is *not* an algebraic integer.


