Is the ring of formal power series in any sort of completion of $\mathbb Q(x)$? This question should be a real soft-ball for any serious algebraist.
So, $\mathbb Q(x)$ obviously contains all polynomials, but also things like $1/(1 - x)$. Now my question is this: Can we make sense of the equation
$$
\sum_{n \ge 0} x^n = \frac{1}{1 - x}
$$
in any extension field of $\mathbb Q(x)$, for instance one that contains $\mathbb Q[[x]]$? What about $\mathbb R$ instead of $\mathbb Q$?
The trouble is that the fraction field of $\mathbb Q[[x]]$ (EDIT: does not seem to make said identity valid is actually the solution, see the answer below). Now one could try to argue from the "real" theorem by identifying expressions that on some open subset of $\mathbb R$ yield the same value, but this approach would fail over different fields where there is no obvious embedding into $\mathbb R$.
Also, in the surreal numbers (or larger ordered fields contained therein), the statement does not quite seem true, because there the notion of convergence is incredibly strong.
 A: What you are after is the field of formal Laurent series over $\mathbb Q$, usually denoted $\mathbb Q((x))$. It is the fraction field of $\mathbb Q[[x]]$, and contrary to what you mention, the identity $\sum_{n\geq 0}x^n=\frac{1}{1-x}$ does hold there - indeed, this simply amounts to checking that $(1-x)\sum_{n\geq 0}x^n=1$, which you can do by considering each coefficient.
Relatedly to the question in the title, $\mathbb Q((x))$ actually is a completion of $\mathbb Q(x)$ in the same way as $\mathbb Q[[x]]$ is a completion of $\mathbb Q[x]$. The metric is defined as follows: for any two distinct $f,g\in\mathbb Q(x)$, write $f-g=x^n\frac{p}{q}$ with $p,q$ polynomials not divisible by $x$, and define $d(f,g)=2^{-n}$. This construction is very similar to the construction of $p$-adic numbers (and, indeed, we can call $\mathbb Q((x))$ the $x$-adic completion of $\mathbb Q(x)$.
Lastly, regarding surreal numbers - you are correct the identity is not true, but this is because the sum in question is not convergent for any $x\neq 0$ (indeed, no sequence of surreals converges unless it is eventually constant.)
