= Solution
We prove convergence of the actual sequential updates above, not just concentration of ideal equilibrium distributions. Write $M_m=\mathbb E v_m$ and $N_m=\mathbb E v_m^2$. Given $v_{m-1}$, the centered normal draw satisfies
$$
\mathbb E[(\mu_m-\bar x)^2\mid v_{m-1}]=\frac{T_m v_{m-1}}n,\qquad
\mathbb E[(\mu_m-\bar x)^4\mid v_{m-1}]=3\left(\frac{T_m v_{m-1}}n\right)^2.
$$
The conditional <inverse-gamma distribution> moments therefore give
$$
M_m=\frac{S+T_mM_{m-1}}{n-4T_m},\qquad
N_m=\frac{S^2+2ST_mM_{m-1}+3T_m^2N_{m-1}}{(n-4T_m)(n-6T_m)}.
$$
For $T_m\leq n/10$, the coefficient multiplying $M_{m-1}$ is at most $1/6$, so $M_m$ stays bounded by a geometric-recursion argument. The coefficient multiplying $N_{m-1}$ is at most $1/8$, and the other terms are bounded using the first-moment bound, so $N_m$ also stays bounded. These bounds start from the finite deterministic $v_0$.
Since $T_m\to0$, the boundedness makes every term containing $T_mM_{m-1}$ or $T_m^2N_{m-1}$ vanish. Hence $M_m\to S/n$ and $N_m\to S^2/n^2$. Consequently
$$
\mathbb E\left(v_m-\frac Sn\right)^2\to0,\qquad
\mathbb E(\mu_m-\bar x)^2=\frac{T_mM_{m-1}}n\to0.
$$
Thus
$$
\boxed{(\mu_m,v_m)\longrightarrow(\bar x,S/n)\quad\text{in mean square and therefore in probability}.}
$$
This establishes <mean-square convergence> for the specified Gaussian annealer. It does not assert that every implementation or cooling rule for every objective has the same guarantee, nor that mean-square convergence alone is almost-sure convergence.
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