Average value of $\frac{x'A^2x}{x'A^3x}$ over surface of $n$-dimensional sphere Suppose $A$ is a diagonal matrix with eigenvalues $1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{n}$ and $x$ is drawn from standard Gaussian in $n$ dimensions. In numerical simulations, the following quantity seems to converge to $2$ as $n\rightarrow \infty$
$$z_n=E_{x\sim \mathcal{N}\left(0, I_n\right)}\left[\frac{x^T A^2 x}{x^T A^3 x}\right]$$
Can this be proven or disproven?
$z_n$ can also be written as the following sum
$$z_n=\sum_{i=1}^n i E_{y\sim \mathcal{N}\left(0,A^3\right)}\left[\frac{y_i^2}{\|y\|^2}\right]$$
This quantity can be viewed as the average ratio of quadratic forms $A^2$ and $A^3$ on the surface of $n$-dimensional sphere.

Here's what the distribution looks like for a few values of $n$, means are tending towards $2$


Not an answer, but wanted to point out an interesting result leading to an identity with the zeta functions. Let $A = \sum_{k\ge 1} X_k^2/k^2$ and $B = \sum_{k \ge 1}X_k^2/k^3$. Then a Taylor series expansion of $f(A, B)$ around $(\mu_A, \mu_B)=\mathbb{E}[(A, B)] = (\zeta(2), \zeta(3))$ is
\begin{align*}
\frac{A}{B} = \frac{\mu_A}{\mu_B} + \begin{pmatrix} 1/\mu_B \\ -\mu_A/\mu_B^2\end{pmatrix}^\intercal \begin{pmatrix} A - \mu_A \\ B - \mu_B \end{pmatrix} + \frac{1}{2}\begin{pmatrix} A - \mu_A \\ B - \mu_B \end{pmatrix}^\intercal\begin{pmatrix} 0 & -1/\mu_B^2 \\ -1/\mu_B^2 & 2 \mu_A/\mu_B^3 \end{pmatrix}\begin{pmatrix} A - \mu_A \\ B - \mu_B \end{pmatrix} + \cdots
\end{align*}
The first few terms of this expansion are
\begin{align*}
\mathbb{E}\left[\frac{A}{B}\right] &= \frac{\mu_A}{\mu_B} - \frac{\mathbb{E}(A - \mu_A)(B-\mu_B)}{\mu_B^2} + \frac{\mu_A\mathbb{E}(B-\mu_B)^2}{\mu_B^3} + \frac{\mathbb{E}(A-\mu_A)^2(B-\mu_B)}{6\mu_B^3} + \frac{\mathbb{E}(A-\mu_A)(B-\mu_B)^2}{3\mu_B^2} - \frac{\mathbb{E}(B-\mu_B)^3\mu_A}{2\mu_B^4} + \cdots \\
&= \frac{\zeta(2)}{\zeta(3)} - 2 \frac{\zeta(5)}{\zeta(3)^2} + 2 \frac{\zeta(6)\zeta(2)}{\zeta(3)^3} + \cdots
\end{align*}
