We just learned about infinite products in class. There's no textbook for the course so I am struggling with the following two basic problems.

Let $a_n(z) = 1 + b_n(z), |b_n(z)| \leq \lambda < 1, z \in E$. Prove that

(1) $\prod (1+|b_n|)$ converges uniformly on $E$ if and only if $\sum |b_n|$ does;

(2) If $\prod (1+|b_n|)$ converges uniformly on $E$, then $\prod (1+b_n)$ does too.

Part (1) seems to be just taking logarithm, because $\prod (1+|b_n|)$ converges uniformly is the equivalent to $\sum \log (1+|b_n|)$ converges uniformly, but $\log (1+|b_n|) \sim |b_n|$ as $b_n \to 0$. Is this correct or does it work only for pointwise convergence? I am not sure how to do part (2).

Any help is appreciated.

Hint for (1):

Using $1 + x \leqslant e^x$ for $x > 0$ we have

$$\sum_{k = n+1}^m |b_k(z)| \leqslant \prod_{ k = n+1}^m (1 + |b_k(z)|) \leqslant \exp \left(\sum_{k = n+1}^m |b_k(z)| \right).$$

Note that by the Cauchy criterion $\sum |b_k(z)|$ converges uniformly if and only if for any $\epsilon > 0$ there exist $N(\epsilon) \in \mathbb{N}$ such for all $m \geqslant n \geqslant N(\epsilon)$ and all $z \in E$ we have

$$\sum_{k = n+1}^m |b_k(z)| < \epsilon$$

Part (2) is a bit tricky.

Let

$$P_n = \prod_{k=1}^n [1 + b_n(z)], \\ R_n = \prod_{k=1}^n [1 + |b_n(z)|].$$

Then

$$|P_n - P_{n-1}| = |(1 +b_1(z)) \ldots (1 + b_{n-1}(z))b_n(z)| \leqslant (1 +|b_1(z)|) \ldots (1 + |b_{n-1}(z)|)|b_n(z)| = R_n - R_{n-1}.$$

If $R_n$ is uniformly convergent, then so too is the telescoping sum $\sum (R_n - R_{n-1})$. By the comparison test, the telescoping sum $\sum (P_n - P_{n-1})$ is unformly convergent. Hence $P_n$ is uniformly convergent.