QM: Time evolution with $H = H(t)$ In order to calculate time evolution in QM we use Schrödinger equation
\begin{align*}
i \partial_t |\psi\rangle_t = H(t) | \psi\rangle_t.
\end{align*}
If $H\neq H(t)$ then 
\begin{align*}
i \partial_t |\psi\rangle_t = H(0) | \psi \rangle_t 
\end{align*}
and we can expand the state in its Taylor series
\begin{align*}
| \psi  \rangle_t & = |\psi\rangle_0 + t \, \partial_t |\psi\rangle_t \Big|_{t=0} + \frac{1}{2} \,  t^2 \, \partial_t^2 |\psi\rangle_t \Big|_{t=0}  + ... \\
& = |\psi\rangle_0 + (-i t H(0)) | \psi \rangle_t \Big|_{t=0} + \frac{1}{2} (-itH(0))^2 | \psi \rangle_t \Big|_{t=0} + ...\\
& = |\psi\rangle_0 + (-i t H(0)) | \psi \rangle_0  + \frac{1}{2} (-itH(0))^2 | \psi \rangle_0  + ...\\
& = e^{-itH(0)}| \psi \rangle_0.
\end{align*}
So far so good. But now we consider $H=H(t)$. My question is: why can't you do the same? Even if now you have $i \partial_t |\psi\rangle_t = H(t) | \psi \rangle_t$ instead of $i \partial_t |\psi\rangle_t = H(0) | \psi \rangle_t$, you still have
\begin{align*}
\partial_t |\psi\rangle_t \Big|_{t=0} = (-i H(t)) |\psi\rangle_t \Big|_{t=0} = (-iH(0)) |\psi\rangle_0.
\end{align*}
According to this you would always get the same time evolution operator:
\begin{align*}
| \psi  \rangle_t & =  e^{-itH(0)}| \psi \rangle_0,
\end{align*}
both for time independent and time dependent operator. 

Of course I realize this doesn't make sense for $H=H(t)$, because it implies that the state at any point is only given by the state and the Hamiltonian at $t=0$, and according to Schrödinger's equation the Hamiltonian "drives" the state at each time. So I just want to know why you can't expand the state in a Taylor series for $H=H(t)$. My guess is that, for some reason, in an isolated system the state is "analytic" and equal to its Taylor series, while for a non isolated system the Taylor series only converges in a neighbourhood of the point, and the correct formula is
\begin{align*}
| \psi  \rangle_{t+\Delta t} & =  e^{-itH(t)}| \psi \rangle_t + \mathcal{O}( \Delta t^2),
\end{align*}
which leads to the general time evolution operator. 
Or maybe it has nothing to do with "analyticity" and it's just somehting silly I'm not seeing right now.
 A: The reason why your argument doesn't work for time-dependent Hamiltonian is that
$$ \left. \partial_t^2 |\psi(t)\rangle \right|_{t=0} = \left. \partial_t (\partial_t |\psi(t)\rangle) \right|_{t=0} = \left. \partial_t (-\mathrm i H(t) |\psi(t)\rangle) \right|_{t=0} \neq \left. (-\mathrm i H(t))^2 |\psi(t)\rangle \right|_{t=0} $$
The time evolution is still analytic, as long as the function $H(t)$ is.
The correct way to do it instead is using a time-ordered exponential
$$ |\psi(t)\rangle = \mathbf{T} \mathrm e^{-\mathrm i \int_{t_0}^t H(\tau)\, \mathrm d\tau} |\psi(t_0)\rangle , $$
which is defined via the Dyson series.
(From your question, I assume that you know this already, so I won't write more about it -- feel free to ask if you have more questions!)
A: The main reason why you can't do those sorts of naïve manipulations is that in general, if $H = H(t)$, hamiltonians evaluated at different times may not commute. If the hamiltonian is time-dependent but $[H(t_1), H(t_2)] = 0$ for all $t_1$ and $t_2$, then you could still analitically solve Schrödinger equation to give
$$|ψ⟩_t=e^{−i\int_0^tdt'H(t')/\hbar}|ψ⟩_0.$$
When hamiltonians at different times don't commute, however, you need to introduce the Dyson time-ordering operator.
