$\DeclareMathOperator{\tr}{tr}$
Consider a Hermitian matrix $H_\lambda = H_0 + \lambda V$, where $H_0, V$ are both Hermitian, and suppose that $H_0$ has an eigenvalue $E_0$. We can choose a sufficiently small contour $\Gamma$ of $E_0$ in $\mathbb{C}$ so that the contour only contains $E_0$ and none of the other eigenvalues of $H_0$. We can then define
$$
P_\lambda = \oint_\Gamma \frac{1}{z-H_\lambda} \frac{dz}{2\pi i}
$$
For sufficiently small $\lambda$, the eigenvalues of $H_\lambda$ only change a little bit with respect to those of $H_0$ (Weyl inequality), and thus it's clear by Cauchy Residues that $P_\lambda$ is the projection operator onto the "perturbed'' eigenspaces of $H_\lambda$, and thus it's easy to see that $\lambda \mapsto P_\lambda$ is analytic. Let's then say we wish to compute the following in orders of $\lambda$
$$
\tr{H_\lambda P_\lambda}
$$
To 2nd order in $\lambda$, we would have the following terms in the expansion
$$
\frac{1}{2} H_0 P_0'' + V P_0'
$$
However, if we wish to instead expand the following in orders of $\lambda$,
$$
\tr{P_\lambda H_\lambda P_\lambda}
$$
(Which should be exactly the same as before due to $P_\lambda$ being a projection operator), we would instead have an additional term of the form
$$
\tr P'_0 H_0 P'_0
$$
(And some other terms that don't really matter). It's not hard to check that this term is nonzero and thus the expansion seems to be ambiguous in some sense?. Indeed, one can check that
$$
P_0' = P_0 V R_0 +R_0 V P_0, \quad R_0 \equiv \frac{1}{E_0 -H_0} (1-P_0)
$$
EDIT. $P_0',P_0''$ are first, second derivative of $\lambda \mapsto P_\lambda$ and taking $\lambda=0$.