Generalized delta method Suppose $f\in C^{k+1}$ satisfies $f^{(j)}(\mu)=0$ for all $1\le j\le k$ and $f^{(j)}(\mu)\ne 0$ where $\mu$ is fixed. Furthermore, assume $a_n\to \infty$ and $a_n(X_n-\mu)$ converges in distribution to some random variable $X$.
I am trying to see if the following two claims are true, which are somewhat generalizations of the Delta method. That is,
$a_n^j(f(X_n)-f(\mu))$ converges in distribution to 0 for all $j$ and that $a_n^{k+1}(f(X_n)-f(\mu))$ converges in distribution to a non-degenerate random variable. However I cannot just mimic the standard proof of the Delta method as it hinges on assuming $a_n=n$.
 A: Let $f \in C^{k+1}$ and let $j \in \{1,...,k\}$ be fixed.
Note that $f(X_n) = f(\mu) + (X_n-\mu)f'(\mu) + ... + \frac{1}{j!}(X_n-\mu)^j f^{(j)}(\mu) + h(X_n-\mu)(X_n-\mu)^j $ where $h$ is such a function, that $h(x) \to 0$ as $x \to 0$.
Note that all derivatives disappear, hence $f(X_n)-f(\mu) = h(X_n-\mu) (X_n-\mu)^j $
Now, the point is, since $a_n(X_n-\mu) \to X$ in distribution, and $a_n \to \infty$, then $(X_n-\mu) \to 0$ in distribution (since $(X_n-\mu) = \frac{1}{a_n} \cdot (a_n(X_n-\mu))$), so in probability, too.
From this follows $h(X_n-\mu) \to 0$ in probability (easiest argument would be to argue by characterisation $Y_n \to Y$ in probability, iff for all $(n_k)$ there exists $(n_{k_m})$ such that $Y_{n_{k_m}} \to Y$ almost surely). By convergence in probability, we have convergence in distribution, too.
Hence $a_n^j(f(X_n)-f(\mu)) = (a_n (X_n-\mu))^j \cdot h(X_n-\mu) \to X^j \cdot 0 = 0$ in distribution (we used that continuous functions preserve convergence in distribution and used slutsky lemma, which says that $Y_n \to Y$ in distribution, and $Z_n \to c$ in distribution, where $c$ is a constant, implies $Y_nZ_n \to cY$ in distribution).
For the last part, do the same writting with remainder, but now for $j=k+1$, we'll get
$f(X_n)-f(\mu) = \frac{1}{(k+1)!}(X_n-\mu)^{k+1}f^{(k+1)}(\mu) + h(X_n-\mu)(X_n-\mu)^{k+1}$
We show as before, that $a_n^{k+1}(X_n-\mu)^{k+1}h(X_n-\mu) \to 0$ in distribution.
And the first part, similarly $(a_n (X_n-\mu))^{k+1} \frac{f^{(k+1)}(\mu)}{(k+1)!} \to X^{k+1} \frac{f^{(k+1)}(\mu)}{(k+1)!}$ in distribution and again slutsky lemma (but now version $Y_n \to Y$ in distribution, $Z_n \to c$ in distribution ($c$ constant) implies $Y_n + Z_n \to Y + c$) and we finally get
$$ a_n^{k+1}(f(X_n) - f(\mu)) \to X^{k+1} \frac{f^{(k+1)}(\mu)}{(k+1)!} $$ in distribution
