Is $\mathbb{Q}[[v^{-1}]] \cap \mathbb{Q}(v)=\mathbb{Q}[[v^{-1}]]$? I have a question about the some rings and fields. Is $\mathbb{Q}[[v^{-1}]] \cap \mathbb{Q}(v)=\mathbb{Q}[[v^{-1}]]$?
 A: To amplify Qiaochu's comment: when you expand a rational function into a power series, you get an eventually periodic sequence of coefficients, basically for the very same reason the decimal expansion of a rational number is eventually periodic. Just use long division. Therefore, there are many series in $\mathbb{Q}[[v^{-1}]]$ that are not in $\mathbb{Q}(v)$. 
EDIT: this was a bit too quick and a bit too simplistic. See David Speyer's comment below. (I was thinking about generating functions over finite fields). 
A: Or how about this: $\mathbb{Q}(v)$ is countable and $\mathbb{Q}[[v^{-1}]]$ isn't. 
Or how about this: take $e^{v^{-1}} = \sum_{n \ge 0} \frac{v^{-n}}{n!}$. The derivatives of $p(v) e^{v^{-1}}$ never vanish identically (induction), so it can never be a polynomial. Hence $e^{v^{-1}}$ can't be a rational function. 
Or, for a counterexample that works in any characteristic, take $f(v) = \sum_{n \ge 0} v^{-n^2}$. If $p(v)$ is a polynomial of degree $d$, then $p(v) f(v)$ has infinitely many nonzero coefficients because the gaps between nonzero terms in $f$ eventually become longer than $d$, so again $f(v)$ can't be a rational function. 
