Equivalent condition for convergent partial products. Is the folllowing argument correct? All cited theorems are presented below.> Proposition. Show, in general, that the sequence of partial products of the following series converges if and only if $\sum\limits_{n=1}^{\infty}a_n$ converges. (The inequality $1 + x\leq 3^x$ for positive $x$ will be useful in one direction.)$$\prod_{n=1}^{\infty}(1+a_n) = (1+a_1)(1+a_2)(1+a_3)\cdots,\quad\text{where }a_n\ge 0.$$Proof. $(\Rightarrow)$. Assume that the sequence $s_n = \prod\limits_{k=1}^{n}(1+a_k)$ of partial products converges, then by theorem 2.3.2 there exists an $M>0$ such that $s_n\leq M,$ $\forall n\in\mathbf{N}$. We now show that $$\sum_{k=1}^{n}a_k\leq \prod_{k=1}^{n}(1+a_k).\quad \forall n\in\mathbf{N}$$The base case is trivial. Now let $k\in\mathbf{N}$ and assume that $\sum\limits_{j=1}^{k}a_j\leq \prod\limits_{j=1}^{k}(1+a_j)$, consequently $$\sum_{j=1}^{k+1}a_j = \sum_{j=1}^{k}a_j +a_{k+1}\leq \prod_{j=1}^{k}(1+a_j)+a_{k+1}.$$ Now assume that $a_{k+1}\left(1-\prod\limits_{j=1}^{k}(1+a_j)\right)>0$, from hypothesis $a_{k+1}\ge 0$, but then $\prod_{j=1}^{k}(1+a_j)< 1$, a contradiction since by hypothesis $(1+a_n)\ge 1$, $\forall n\in\mathbf{N}$, thus $a_{k+1}\leq a_{k+1}\prod\limits_{j=1}^{k}(1+a_j)$ and then by inductive hypothesis we have\begin{align*}\sum_{j=1}^{k+1}a_j &= \sum_{j=1}^{k}a_j +a_{k+1} \leq\prod_{j=1}^{k}(1+a_j)+a_{k+1}\\&\leq \prod_{j=1}^{k}(1+a_j)+ a_{k+1}\prod_{j=1}^{k}(1+a_j) = \prod_{j=1}^{k+1}(1+a_j),\end{align*}completing the induction. It is now apparent that $\sum\limits_{k=1}^{n}a_k\leq M$, $\forall n\in\mathbf{N}$, and since $a_n\ge 0$ it follows that the corresponding sequence of partial sums is increasing, the series $\sum_{k=1}^{\infty}a_k$ is then convergent by theorem 2.4.1.$(\Leftarrow)$. Now assume that the sequence $b_n = \sum\limits_{j=1}^{n}a_j$ of partial sums is convergent, again by theorem 2.3.2 there exists a $M>0$ such that $b_n\leq M$, $\forall n\in\mathbf{N}$. We know that $(1+x)<3^x$ when $x\ge 0$ consequently $$\prod_{j=1}^{k}(1+a_j)<3^{\sum_{j=1}^{k}a_j}\leq 3^M.$$In summary then $\prod\limits_{k=1}^{n}(1+a_k)\leq 3^M$, $\forall n\in\mathbf{N}$, again since $a_n\ge 0$, $\forall n\in\mathbf{N}$ it follows that corresponding sequence of partial products is increasing, appealing to theorem 2.4.1 implies that $\prod\limits_{n=1}^{\infty}(1+a_n)$ is convergent.$\blacksquare$----Note:  - Theorem (2.4.1): Every bounded montonic sequence converges.  - Theorem (2.3.2): Every convergent sequence is bounded.
 A: ﻿Your proof is good, but there are still a few points to note.For the necessity part, what you proved is a weakened generalization of Bernoulli's inequality. A simpler inductive proof can prove the following generalization:>If $a_1, \cdots, a_n \geqslant 0$, then$$\prod_{k = 1}^n (1 + a_k) \geqslant 1 + \sum_{k = 1}^n a_k.$$Proof: For $n = 1$, it is trivial. For $n = 2$, since $a_1, a_2 \geqslant 0$, then$$(1 + a_1)(1 + a_2) = 1 + a_1 + a_2 + a_1 a_2 \geqslant 1 + a_1 + a_2.$$Now assume that it holds for $n$ ($n \geqslant 2$), then\begin{align*}&\mathrel{\phantom{=}}{} \prod_{k = 1}^{n + 1} (1 + a_k) = \left( \prod_{k = 1}^n (1 + a_k) \right) (1 + a_{n + 1})\\&\geqslant \left( 1 + \sum_{k = 1}^n a_k \right)(1 + a_{n + 1}) \geqslant 1 + \sum_{k = 1}^n a_k + a_{n + 1} = 1 + \sum_{k = 1}^{n + 1} a_k.\end{align*}End of induction.For the sufficiency part, the convergence of an infinite product usually requires that the limit of the sequence of partial products is not $0$. Here it only has to be notified that $\prod\limits_{k = 1}^n (1 + a_k) \geqslant 1 > 0$ for all $n$. And incidentally, the inequality $1 + x \leqslant 3^x$ for $x \geqslant 0$ can be replaced by a sharper inequality, i.e. $1 + x \leqslant \mathrm{e}^x$.
