Let $K = \bigcup_{p\ \nmid\ n}\Bbb{Q}_p(\zeta_n)$ the maximal unramified extension of $\Bbb{Q}_p$. Then $K$ is smaller than its completion $K_v$: $$a=\sum_{n=1,p\ \nmid\ n}^\infty \zeta_n p^n\in K_v$$ $a$ is not algebraic over $\Bbb{Q}_p$: the Frobenius $\phi(\zeta_n)=\zeta_n^p$ is in $Aut(K_v/\Bbb{Q}_p)$, it sends $K$ to itself, and $\forall r,\phi^r(a) \ne a$.
From this action of the Frobenius we see $K=K_v\cap \overline{\Bbb{Q}}_p$ is the subfield of $K_v$ of p-adic series $\sum_n c_n p^n$ with the non-zero $c_n$ taken in a set of finitely many roots of unity.
In particular if $f\in \Bbb{Z}_p[x]$ is monic and separable modulo $p$ then it splits completely in $K_v$ thus in $K$ which wasn't quite obvious at first.
$\Bbb{Q}_p(a)/\Bbb{Q}_p$ is a non-algebraic extension and its completion is the complete DVR $K_v$. Moreover it is the completion, not $K$ which appears from $\varprojlim \Bbb{Z}[\zeta_\infty]/\mathfrak{p}^m$ for $\mathfrak{p}$ a prime ideal above $p$.
I wonder if $K_v$ as well as the completion of $\bigcup_{p\ \nmid\ n} K(p^{1/n})$ have simpler properties than the non-complete version in particular in term of their wildly ramified extensions ?