Assume we have hamiltonian
$$ H = H_0 + \lambda V$$
where $ H_0 $ is unperturbed hamiltonian which we know the eigenstastes, and $ V$ is a perturbation.
In the effective hamiltonian approach using the canonical transformation, we transform the hamiltonian via
$$ H_{eff} = e^{S}He^{-S}$$
where $ S^\dagger = -S$, so $e^S$ is hermitian operator, so we are actually doing an unitary transformation of hamiltonian. Then expanding this term using the identity
$$ e^{S}He^{-S} = H + [S,H] + \frac{1}{2}[S,[S,H]] + \frac{1}{3!}[S,[S,[S,H]]] + ...$$
we get the effective hamiltonian as
$$ H_{eff} = H_0 + \lambda V + [S,H_0] + \lambda[S,V] + \frac{1}{2}[S,[S,H_0]] + \lambda \frac{1}{2}[S,[S,V]] + ...$$
But after the transformation, above hamiltonian looks more complicated than the original one. In this step, I don't see any reason why we do the canonical transformation to get the effective hamiltonian.
The book says that if we can find the operator $S$ which satisfies
$$\lambda V + [S,H_0] = 0$$
then we can eliminate the second and third terms of effective hamiltonian, but we still have the infinite series terms in the hamiltonian, which still looks complicated.
What is the essential point of this canonical transformion approach when getting the effective hamiltonian?