 Solution

For $x\ge 0$, the function $h(x)=e^x-1-x$ has $h(0)=0$ and $h^{\prime }(x)=e^x-1\ge 0$, because the exponential function is increasing and $e^0=1$. Therefore $\boxed{e^x\ge 1+x}$.

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 $P_n\ge 1+S_n\ge S_n$. Applying the exponential inequality to every factor givesBoth sequences are increasing, since the terms $a_j$ are positive. If $S_n$ converges to a finite value $S$, then $P_n\le e^S$, so the convergence theorem for monotone bounded sequences makes $P_n$ converge. Conversely, if $P_n$ has a finite limit, then $S_n\le P_n$ is bounded and increasing, hence converges by the same theorem. Thus 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.