= Solution
With the natural <filtration> $\mathcal F_n=\sigma(Z_1,\ldots,Z_n)$, the sums form a <martingale>. Put $v_n=\sum_{k\leq n}a_k^2$. <Independence> and the <characteristic function> of a <standard normal random variable> give
$$
\mathbb E e^{iuM_n}=e^{-u^2v_n/2}.
$$
If $M_n\to M_\infty$ <almost surely> with a finite limit, <dominated convergence> gives convergence of these <characteristic functions> to $\mathbb E e^{iuM_\infty}$. If $v_n\to\infty$, that limit would be $1$ at zero and $0$ at every nonzero $u$. A <characteristic function> is continuous at zero: apply <dominated convergence> to $e^{iuM_\infty}$ as $u\to0$. This contradiction proves
$$
\boxed{\sum_{k=1}^\infty a_k^2<\infty.}
$$
This is also sufficient by the <L2-bounded martingale convergence theorem>, since $\mathbb EM_n^2=v_n$. Thus the <almost sure convergence criterion for an independent Gaussian series> is an equivalence.
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