finitely tailed Laurent series as the completion of $k(T)$ I've been trying to solve the following problem:
Show that $k((1/T))$, the ring of finitely tailed Laurent series in $1/T$ with coefficients in $k$, is the completion of $k(T)$ with respect to a discrete absolute value associated with the valuation $v_\infty$ of $k(T)$.
I have no idea about how to relate the aforementioned expansion with the formal construction of completion (taking Cauchy sequences and doing quotient by the ones approaching $0$). Hints?
 A: Note: I assume here the (additive, nonarchimedean) valuation $v_\infty$ on $k(T)$ is given by $v_\infty \left(\dfrac{p(T)}{q(T)} \right) := \deg(q)-\deg(p)$ as it should be.
$\displaystyle\dfrac{1}{c_0+c_1 T+ ... +c_n T^n}= \frac{1}{c_n}\cdot T^{-n} \cdot \dfrac{1}{\frac{c_0}{c_n} (1/T)^n+\frac{c_1}{c_n}(1/T)^{n-1}+...+1} \\ \displaystyle \stackrel{\text{geometric series}}= \frac{1}{c_n} \cdot (1/T)^{n} \cdot \left( \sum_{k=0}^\infty \left(-\frac{c_0}{c_n} (1/T)^n-\frac{c_1}{c_n}(1/T)^{n-1}-...-\frac{c_{n-1}}{c_n}(1/T)\right)^k \right)$
and this clearly is an element of $k((1/T))$ of valuation $n$ (where the additive valuation $v_{1/T}$ is just the lowest index of the Laurent series in the variable $(1/T)$ with non-zero coefficient). Hence a general element of $k(T)$,
$\displaystyle\dfrac{a_0+a_1 T +...+a_mT^m}{c_0+c_1 T+ ... +c_n T^n} \\ \displaystyle =a_0 \cdot \dfrac{1}{c_0+c_1 T+ ... +c_n T^n} + a_1 T \cdot \dfrac{1}{c_0+c_1 T+ ... +c_n T^n}+ ...+ a_m T^m \cdot \dfrac{1}{c_0+c_1 T+ ... +c_n T^n} \\ \displaystyle =a_0 \cdot \dfrac{1}{c_0+c_1 T+ ... +c_n T^n} + a_1 (1/T)^{-1} \cdot \dfrac{1}{c_0+c_1 T+ ... c_n T^n}+ ...+ a_m (1/T)^{-m} \cdot \dfrac{1}{c_0+c_1 T+ ... +c_n T^n}$
is an element of $k((1/T))$ of valuation $n-m$.
Formally, the above defines an isometric embedding of rings
$$(k(T), v_{\infty}) \hookrightarrow (k((1/T)), v_{1/T})$$
and since the latter is very well-known to be complete, all that is left to show is that the image of the former is dense in it. Well, for example it clearly contains all Laurent polynomials.
