Suppose $A$ is a $d\times d$ positive definite convergent matrix and $t>1$. How do I upper bound the following in terms of simple statistics of $A$?
$$f_A(t)=\operatorname{Tr}\exp(-At)-\operatorname{Tr}(I-A)^t$$
In particular I'm wondering if $f_A(t)\to 0$ as we consider large "heavy-tailed matrices. Concretely, as $d\to \infty$
$$ \operatorname{Tr}(A_d)\to \infty, \operatorname{Tr}(A_d^2)\to 0\\ \text{implies}\\ f_{A_d}(t)\to 0 $$
Example, plotting $f_A(t)$ for diagonal $A$ with diagonal values $1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{100}$
Motivation: this gives the error of using continuous time approximation to model $x_{t+1}=(I-A)x_t$ discrete linear system.
We may assume without losing generality, we may assume $A$ is diagonal. Then we have$f_A(t) = \sum_{i=1}^d \{e^{-A_{ii} t} - (1-A_{ii})^t \}$.
We show the following inequality, $$ -\frac{t(t-1)}{2} \lambda^2 \leq e^{-\lambda t} -(1-\lambda)^t \leq \frac{t^2}{2}\lambda^2. $$
First note that $1-x \leq e^{-x} \leq 1 - x + \frac{x^2}{2}$ holds for any $x > 0$. Also note that $1 - t\lambda \leq (1-\lambda)^t \leq 1 - t\lambda + \frac{t(t-1)}{2}\lambda^2$ holds for $0<\lambda<1$. Both these inequalities can be verified easily by taking derivatives up to second order. Combining these, we obtain the inequality above.
From the inequality, we conclude
$$ -\frac{t(t-1)}{2} \text{Tr}(A_d^2) \leq f_A(t) \leq \frac{t^2}{2}\text{Tr}(A_d^2) $$ for sufficiently large $d$.
So the assumption $\text{Tr}(A_d^2)\to 0$ implies your claim.