 Solution

Taking logarithms of the product givesBy part (d)(i), $V_j=E_j-\log b$ in probability distribution, where $E_j$ has the exponential distribution of rate one. Thus $\mathbb{E}|V_j|\le 1+|\log b|<\infty$ and $\mathbb{E}{V}_j=1-\log b$. The strong law of large numbers givesIf $b<e$, choose $\delta =(1-\log b)/2>0$. Eventually $\log W_n\le -\delta n$, so $W_n\le e^{-\delta n}\to 0$. If $b>e$, choose $\delta =(\log b-1)/2>0$; eventually $W_n\ge e^{\delta n}\to \infty$. ThereforeIn 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.