Can we obtain an exact answer when we sum Feynman diagrams to all orders? Consider a $\phi^4$ theory in QFT. Following Peskin & Schroeder's QFT chapter 4, we can do some calculations of correlation functions using perturbation expansion. On their book's page 83, they consider the interaction picture field $\phi_I(t,\textbf{x})$
$$
\phi_I(t, \mathbf{x})=\left.\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_{\mathbf{p}}}}\left(a_{\mathbf{p}} e^{-i p \cdot x}+a_{\mathbf{p}}^{\dagger} e^{i p \cdot x}\right)\right|_{x^0=t-t_0} \tag{4.15}
$$
then
$$
H_I(t)=e^{i H_0\left(t-t_0\right)}\left(H_{\mathrm{int}}\right) e^{-i H_0\left(t-t_0\right)}=\int d^3 x \frac{\lambda}{4 !} \phi_I^4 \tag{4.19}
$$
So (4.15) is very important. And $\text{exp}[-i\int_T^T dt  H_I(t)]$ is a crucial part to calculate the correlation function. For example, (4.31)
$$
\langle\Omega|T\{\phi(x) \phi(y)\}| \Omega\rangle=\lim _{T \rightarrow \infty(1-i \epsilon)} \frac{\left\langle 0\left|T\left\{\phi_I(x) \phi_I(y) \exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle}{\left\langle 0\left|T\left\{\exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle} \tag{4.31}
$$
But, I think (4.15) is an $\textbf{approximation}$ in $\phi^4$ theory. Since (4.15) implies KG equation
$$(\partial^2+m^2)\phi_I(t,\textbf{x})=0 \tag{A}$$
this can be derived from the Heisenberg equation
$$\frac{\partial \phi_I(t,\textbf{x})}{\partial t}=i[H_0,\phi_I(t,\textbf{x})] \tag{B}$$
where $H_0$ is the free field Hamiltonian.
I think (B) is an approximation, since we need to use the full $H$, instead of $H_0$, where $H=H_0+\frac{\lambda \phi^4}{4!}$.
Does this means even though we expend (4.31) to all orders, we still cannot an exact answer, since $\phi_I(t, \mathbf{x})$ is an approximation from the first place?
 A: In the interaction picture, the time evolution of operators is given by the free Hamiltonian
$$
\left(\square + m^2\right) \phi_I = 0
$$
and the time evolution of states is determined by the interaction Hamiltonian
$$
i \frac{\partial |\Psi\rangle_I}{\partial t} = H_I |\Psi\rangle_I
$$
This is not an approximation. The interaction picture is related to the Schrodinger and Heisenberg pictures by unitary transformations. You can describe the dynamics exactly in any of these pictures.
However, summing Feynman diagrams to all orders does not give an exact answer -- but for a different reason than the one you brought up. In particular, the Feynman diagram expansion is an asymptotic expansion. The series actually diverges if you include all terms; for a given value of the coupling, there is an optimal number of terms to include that gets you close to the right answer. The ultimate issue is that the exact transition amplitudes include non-perturbative effects such as instantons that cannot be computed within perturbation theory.
