= Solution
Taking <logarithms> of the product gives
$$
\boxed{W_n=\exp\left(-\sum_{j=1}^nV_j\right).}
$$
By part (d)(i), $V_j=E_j-\log b$ in <probability distribution>, where $E_j$ has the <exponential distribution> of rate one. Thus $\mathbb E|V_j|\leq1+|\log b|<\infty$ and $\mathbb EV_j=1-\log b$. The <strong law of large numbers> gives
$$
\frac1n\log W_n=-\frac1n\sum_{j=1}^nV_j\longrightarrow\log b-1\quad\text{almost surely}.
$$
If $b<e$, choose $\delta=(1-\log b)/2>0$. Eventually $\log W_n\leq-\delta n$, so $W_n\leq e^{-\delta n}\to0$. If $b>e$, choose $\delta=(\log b-1)/2>0$; eventually $W_n\geq e^{\delta n}\to\infty$. Therefore
$$
\boxed{W_n\to0\text{ a.s. if }0<b<e;\qquad W_n\to\infty\text{ a.s. if }b>e.}
$$
In particular, the second range cannot have <almost sure convergence> to zero. This is the <almost-sure growth rate of independent products>. At $b=e$ the logarithmic limit is zero, which by itself decides neither behavior; that boundary case is not requested.
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