Covariant derivative vs partial derivative Question
$\partial_\mu \vec{V} = 0$, does this imply that $\nabla_\mu V^\kappa = 0$ ?
My argument
For any vector $\vec{V}$ we can write that $V=V^\kappa \vec{g_\kappa}$ with $\vec{g_\kappa} = \partial \vec{r}/\partial r^\kappa$ such that:
$$0= \partial_\mu \vec{V} = \partial_\mu(V^\kappa \vec{g}_\kappa) = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\kappa \partial_\mu(\vec{g_\kappa}) = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\kappa \Gamma_{\mu\kappa}^\rho \vec{g_\rho} = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\rho \Gamma_{\mu\rho}^\kappa \vec{g_\kappa} = \nabla_\mu (V^\kappa) \vec{g_\kappa}$$
projecting onto the different basis vectors than gives the result that $\nabla_\mu V^\kappa = 0$
Additional question
If this is true, and we consider a translation in space $\vec{x} \rightarrow \vec{x} + \vec{a}$ with $\vec{a}$ constant. Do we than find that $\nabla_\mu a_\nu = 0$?
 A: The initial problem comes when you start with the quantity $\partial_{\mu} \vec{V}$. This is not a valid operation - remember generally vectors are defined in the tangent space - they are linear differential operators. You can only operate on the components of the vector i.e. $\partial_{\mu} V^{\mu}$.
Instead, for some path in a manifold with coordinates $x^i$ and parameterized by $\lambda$, we can ask how does the geometrical object $\vec{V}$ change along the path?
In this case we have,
\begin{align*}
\frac{d\vec{V}}{d \lambda} = \frac{d}{d \lambda}(V^i e_i) &= \frac{d V^i}{d \lambda} e_i + V^i \frac{d e_i}{d\lambda} \\
&= \frac{d V^i}{d \lambda} e_i + V^i \frac{d e_i}{dx^b} \frac{dx^b}{d \lambda}\\
&=\frac{d V^i}{d \lambda} e_i + V^i \Gamma_{ib}^c e_c \frac{dx^b}{d \lambda}
\end{align*}
If we are just interested in how the components of the vector change, then we can relabel the indices and drop the basis vectors such that
$$ \frac{D V^i}{d \lambda} = \frac{d V^i}{d \lambda} + V^i \Gamma^a_{ic} \frac{d x^c}{d \lambda}$$
