$\text{Prove }\prod_{i=1}^\infty(1+a_i) \text{ converges } \iff \sum_{n=1}^\infty a_n \text{ converges}$ Let $a_i \ge 0$ $$\text{Prove }\prod_{i=1}^\infty(1+a_i) \text{ converges } \iff \sum_{n=1}^\infty a_n \text{ converges}$$
I've got to this step $$\prod_{i=1}^\infty (1+a_i) = e^{\sum_{i=1}^\infty \ln(1+a_i)}$$
but I'm not sure how to proceed
 A: Hint: $1+x\leq e^x$ for all $x\geq 0$. This can be proved using the mean value theorem.
A: First of all
$$\prod_{i=1}^\infty(1+a_i) \text{ converges } \iff \sum_{n=1}^\infty\ln(1+ a_n) \text{ converges}$$
$\forall a_n=0$, we have $\ln(1+a_n)=a_n=0$
$\forall a_n>0$,
$$\prod_{i=1}^\infty(1+a_i) \text{ converges }\Rightarrow \lim_{n\to\infty}a_n=0$$
$$\sum_{n=1}^\infty\ln(1+ a_n) \text{ converges}\Rightarrow\lim_{n\to\infty}a_n=0$$
Thus $a_n\to 0$, which yields
$$\lim_{n\to\infty}\frac{\ln(1+a_n)}{a_n}=1$$
In conclusion, by comparison test
$$\sum_{n=1}^\infty\ln(1+ a_n) \text{ converges}\iff\sum_{n=1}^\infty\ a_n \text{ converges}$$
QED.
A: $\prod_{i=1}^\infty (1 + a_i)$ converges $\iff$ for all $\epsilon \in (0,1/2)$, there exists $N$ such that if $N_1 > N_2 \geq N$ then $\prod_{i=N_1}^{N_2} (1+a_i) \in (1-\epsilon,1+\epsilon)$ $\iff$ for all $\epsilon \in (0,1/2)$, there exists $N$ such that if $N_1 > N_2 \geq N$ then $\sum_{i=N_1}^{N_2} \log(1+a_i) \in (\log(1-\epsilon),\log(1+\epsilon))$.
Furthermore, $\sum_{i=1}^N a_i$ converges $\iff$ for all $\epsilon \in (0,1/2)$, there exists $N$ such that if $N_1 > N_2 \geq N$ then $\sum_{i=N_1}^{N_2} a_i \in (-\epsilon,+\epsilon)$.
The last two statements in each of the preceding paragraphs are equivalent with one another because of the inequality: $\frac{2}{3}x \leq \log(1+x) \leq 2x$ for all $x \in (-1/2,1/2)$.
