= Solution
For $x\geq0$, the function $h(x)=e^x-1-x$ has $h(0)=0$ and $h'(x)=e^x-1\geq0$, because the <exponential function> is increasing and $e^0=1$. Therefore $\boxed{e^x\geq1+x}$.
Write $S_n=\sum_{j=1}^n a_j$ and $P_n=\prod_{j=1}^n(1+a_j)$. Expanding the product produces $1+S_n$ and additional nonnegative terms, giving $P_n\geq1+S_n\geq S_n$. Applying the exponential inequality to every factor gives
$$
\boxed{S_n\leq P_n\leq\prod_{j=1}^n e^{a_j}=e^{S_n}.}
$$
Both sequences are increasing, since the terms $a_j$ are positive. If $S_n$ converges to a finite value $S$, then $P_n\leq e^S$, so the convergence theorem for <monotone bounded sequences> makes $P_n$ converge. Conversely, if $P_n$ has a finite limit, then $S_n\leq P_n$ is bounded and increasing, hence converges by the same theorem. Thus \b[the positive series and the product both converge to finite limits, or both diverge to $+\infty$]. This proves the <positive sum-product convergence criterion>. The word limit here means a finite real limit; allowing extended infinite limits would make the conclusion vacuous for these <monotone sequences>.
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