= Solution
The <autocovariance> vanishes beyond lag $q$. For an ordinary consecutive sum, <variance of a sum> gives
$$
\operatorname{Var}\!\left(\frac1{\sqrt n}\sum_{t=1}^nX_t\right)
=\sum_{|h|<n}\left(1-\frac{|h|}{n}\right)\gamma(h)
\longrightarrow\sum_{h\in\mathbb Z}\gamma(h).
$$
The last sum is finite. For $h\geq0$, $\gamma(h)=\sigma^2\sum_{j=0}^{q-h}\theta_j\theta_{j+h}$. Counting all coefficient pairs gives
$$
\boxed{\lim_{n\to\infty}\operatorname{Var}\!\left(\frac1{\sqrt n}\sum_{t=1}^nX_t\right)
=\sigma^2\left(\sum_{j=0}^q\theta_j\right)^2=\sigma^2\Theta(1)^2.}
$$
For the odd-indexed sample the lag-$h$ <autocovariance> is $\gamma(2h)$, so the same finite-sum argument gives $\sum_h\gamma(2h)$. Only coefficient pairs of the same parity contribute. If $E=\sum_{j\ {\rm even}}\theta_j$ and $O=\sum_{j\ {\rm odd}}\theta_j$, the \b[odd-subsample limit] is
$$
\boxed{\lim_{n\to\infty}\operatorname{Var}\!\left(\frac1{\sqrt n}\sum_{t=1}^nX_{2t-1}\right)
=\sigma^2(E^2+O^2)=\frac{\sigma^2}{2}\bigl[\Theta(1)^2+\Theta(-1)^2\bigr].}
$$
This is the <odd-subsample long-run variance of a moving average>. Equivalently it is $[f_X(0)+f_X(1/2)]/2$ in the cycles convention. Under the printed invertibility assumption and positive noise <variance>, the polynomial is nonzero at both $1$ and $-1$, so both limits are positive. The same formulas also hold without invertibility, when cancellations can make a limit zero.
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