[step:Compute the second-order moments and identify the limiting covariance]Since $\mathbb E[X_t]=0$, multilinearity gives $\mathbb E[d_n(\omega)]=0$, hence every coordinate of $V_n$ has mean $0$ (all first cumulants vanish). For the second order, introduce the second-moment kernel
\begin{align*}
R_n(\alpha,\beta):=\mathbb E\!\big[d_n(\alpha)\,\overline{d_n(\beta)}\big]=\frac{1}{2\pi n}\sum_{s=1}^{n}\sum_{t=1}^{n}\gamma(s-t)\,e^{-i\alpha s+i\beta t},
\end{align*}
where we used $\mathbb E[X_sX_t]=\gamma(s-t)$. Substituting $s=t+h$ and writing $e^{-i\alpha s+i\beta t}=e^{-i\alpha h}e^{-i(\alpha-\beta)t}$,
\begin{align*}
R_n(\alpha,\beta)=\frac{1}{2\pi n}\sum_{|h|<n}\gamma(h)\,e^{-i\alpha h}\sum_{t\in I_h}e^{-i(\alpha-\beta)t},\qquad I_h:=\{t: 1\le t\le n,\ 1\le t+h\le n\},
\end{align*}
where $I_h$ is an integer interval with $|I_h|=n-|h|$. Note $\mathbb E[d_n(\alpha)d_n(\beta)]=R_n(\alpha,-\beta)$.
**Diagonal case $\alpha=\beta=\omega_n\to\omega$.** Then $\alpha-\beta=0$, the inner sum equals $|I_h|=n-|h|$, and
\begin{align*}
R_n(\omega_n,\omega_n)=\frac{1}{2\pi}\sum_{|h|<n}\Big(1-\tfrac{|h|}{n}\Big)\gamma(h)\,e^{-i\omega_n h}.
\end{align*}
The summand, viewed as a function on $\mathbb Z$ with counting measure, is dominated by $|\gamma(h)|\in\ell^1(\mathbb Z)$ and converges pointwise to $\gamma(h)e^{-i\omega h}$ as $n\to\infty$ (since $1-|h|/n\to1$ and $e^{-i\omega_n h}\to e^{-i\omega h}$). By the [Dominated Convergence Theorem](/theorems/4) applied to the $\sigma$-finite counting measure on $\mathbb Z$,
\begin{align*}
R_n(\omega_n,\omega_n)\ \longrightarrow\ \frac{1}{2\pi}\sum_{h\in\mathbb Z}\gamma(h)e^{-i\omega h}=f(\omega).
\end{align*}
**Off-diagonal case $\alpha_n-\beta_n\to\rho\not\equiv 0\pmod{2\pi}$.** Then $e^{-i(\alpha_n-\beta_n)}\to e^{-i\rho}\ne 1$, so for all large $n$ we have $|1-e^{-i(\alpha_n-\beta_n)}|\ge\tfrac12|1-e^{-i\rho}|=:\delta>0$. Summing the geometric progression over the interval $I_h$,
\begin{align*}
\Big|\sum_{t\in I_h}e^{-i(\alpha_n-\beta_n)t}\Big|\le\frac{2}{|1-e^{-i(\alpha_n-\beta_n)}|}\le\frac{2}{\delta}\qquad(\text{all large }n),
\end{align*}
a bound independent of $h$ and $n$. Therefore
\begin{align*}
|R_n(\alpha_n,\beta_n)|\le\frac{1}{2\pi n}\cdot\frac{2}{\delta}\sum_{h\in\mathbb Z}|\gamma(h)|=O\!\big(n^{-1}\big)\ \longrightarrow\ 0.
\end{align*}
We now apply these two cases. All the relevant frequency combinations are nonzero modulo $2\pi$ because $\omega_1,\dots,\omega_p$ are distinct points of $(0,\pi)$:
- **Variance.** $R_n(\omega_{r,n},\omega_{r,n})=\mathbb E\big[(A_{r,n})^2+(B_{r,n})^2\big]\to f(\omega_r)$.
- **Complementary covariance.** $\mathbb E[d_n(\omega_{r,n})^2]=R_n(\omega_{r,n},-\omega_{r,n})$ has frequency difference $\alpha_n-\beta_n=2\omega_{r,n}\to 2\omega_r\in(0,2\pi)$, hence $\not\equiv 0\pmod{2\pi}$; so $\mathbb E[d_n(\omega_{r,n})^2]\to 0$. Expanding $d_n^2=(A-iB)^2=A^2-B^2-2iAB$, the real and imaginary parts give
\begin{align*}
\mathbb E\big[(A_{r,n})^2\big]-\mathbb E\big[(B_{r,n})^2\big]\to 0,\qquad \mathbb E\big[A_{r,n}B_{r,n}\big]\to 0.
\end{align*}
Together with the variance limit, this yields
\begin{align*}
\mathbb E\big[(A_{r,n})^2\big]\to\frac{f(\omega_r)}{2},\quad \mathbb E\big[(B_{r,n})^2\big]\to\frac{f(\omega_r)}{2},\quad \mathbb E\big[A_{r,n}B_{r,n}\big]\to 0.
\end{align*}
- **Cross-frequency covariances ($r\ne s$).** Both $R_n(\omega_{r,n},\omega_{s,n})$ (difference $\to\omega_r-\omega_s\in(-\pi,\pi)\setminus\{0\}$) and $R_n(\omega_{r,n},-\omega_{s,n})$ (difference $\to\omega_r+\omega_s\in(0,2\pi)$) have frequency differences $\not\equiv 0\pmod{2\pi}$, hence both tend to $0$. Expanding $\mathbb E[d_n(\omega_{r,n})\overline{d_n(\omega_{s,n})}]$ and $\mathbb E[d_n(\omega_{r,n})d_n(\omega_{s,n})]$ into real and imaginary parts as above shows that all four cross-covariances $\mathbb E[A_{r,n}A_{s,n}],\ \mathbb E[A_{r,n}B_{s,n}],\ \mathbb E[B_{r,n}A_{s,n}],\ \mathbb E[B_{r,n}B_{s,n}]$ tend to $0$.
Hence the covariance matrix of $V_n$ converges to the block-diagonal matrix
\begin{align*}
\Sigma=\operatorname{diag}\!\Big(\tfrac{f(\omega_1)}{2},\tfrac{f(\omega_1)}{2},\ \dots,\ \tfrac{f(\omega_p)}{2},\tfrac{f(\omega_p)}{2}\Big)\in\mathbb R^{2p\times 2p}.
\end{align*}[/step]