I'm trying to derive a very basic result stated in several books on random matrix theory (e.g. Terry Tao's book and Potters & Bouchaud's book).


Given a symmetric matrix $A \in \mathbb{R}^{N \times N}$, with eigenvalues $\lambda\_1, \dots, \lambda\_n$, define a complex-valued function $g : \mathbb{C} \mapsto \mathbb{C}$ as follows:


$$
g(z) = \frac{1}{N} \text{Tr} [ (zI - A)^{-1} ] = \frac{1}{N} \sum\_{n=1}^N \frac{1}{z - \lambda\_n}
$$


This is called the Stieltjes transform of $A$. For sufficiently large $z$, it is claimed that the Taylor expansion can be written as (see eq. 2.22 in Potters & Bouchaud):


$$
g(z) = \sum\_{k=0}^\infty \frac{1}{z^{k + 1}} \frac{1}{N} \text{Tr}[A^k]
$$


Both books emphasize that this works *for sufficiently large* $z$ and I suspect this is the key piece of information I'm missing. Because If I try to expand around $z = 0$, I get:


$$
g(z) = \sum\_{k=0}^\infty \frac{z^k}{k!} \cdot g^{(k)}(0) = \sum\_{k=0}^\infty \frac{z^k}{k!} \sum\_{n=1}^N \frac{(-1)^k \cdot k!}{N \cdot (-\lambda\_n)^k} = \sum\_{k=0}^\infty \frac{z^k}{N} \sum\_{n=1}^N \frac{1}{\lambda\_n^k} = \sum\_{k=0}^\infty z^k \frac{1}{N} \text{Tr}[A^{-k}]
$$


which is more-or-less the reciprocal of what I was hoping for. This is a total stab in the dark, but is there some justification for expanding instead like...


$$
g(z) = \sum\_{k=0}^\infty \frac{k!}{z^k \cdot g^{(k)}(0)}
$$



Taylor expansion of Stieltjes Transform
You certainly can't expand around $z=0$ if "$z$ has to be sufficiently large". Expand around $1/z =0$:
$$
\frac{1}{N} \sum\_{n=1}^{N}\frac{1}{z-\lambda\_{n} } =
\frac{1}{N} \frac{1}{z} \sum\_{n=1}^{N}\frac{1}{1-\lambda\_{n}/z }
=\frac{1}{N} \frac{1}{z} \sum\_{n=1}^{N} \sum\_{k=0}^{\infty } \frac{\lambda\_{n}^{k} }{z^k } = \sum\_{k=0}^{\infty } \frac{1}{z^{k+1} } \frac{1}{N} \mbox{Tr } [A^k ]
$$
(it seems your quoted result has an error in the summation index - it should start at $k=0$, not $k=1$).


