= Solution
The exact condition is again \b[$\sum_ka_k^2<\infty$], with the terminal almost-sure limit used to define $M_T$ when $T=\infty$. In that case $\sup_n\mathbb EM_n^2<\infty$, and
$$
\mathbb E[|M_n|;|M_n|>R]\leq\frac{\mathbb EM_n^2}{R}
$$
proves <uniform integrability>. The preceding unbounded <optional stopping> argument yields $\boxed{\mathbb E M_T=0}$.
For necessity, suppose $v_n\to\infty$. For every fixed $R>0$, the <normal distribution> of $M_n$ gives $\mathbb P(M_n\geq R)\to1/2$. Hence $\mathbb P(\sup_nM_n\geq R)\geq1/2$, and intersecting over positive integer $R$ shows $\mathbb P(\sup_nM_n=\infty)\geq1/2$. Unboundedness above is unchanged by altering finitely many of the <independent> $Z_k$, because that changes all sufficiently late sums by one finite constant. It is therefore a <tail event>; the <Kolmogorov zero-one law> makes its probability one. Consequently $T<\infty$ <almost surely> and $M_T\geq1$. Its <expectation> is at least one, possibly infinite, and cannot equal zero. This proves the necessity assertion in <threshold stopping of an independent Gaussian series>.
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