Let $(a_n)_{n\geq 1}$ be a sequence of complex numbers. The infinite product $\prod_{k=1}^\infty a_k$ is said to converge if for all $\epsilon>0$ there is $N\in\mathbb{N}$ such that $|a_{n+1}\cdots a_{n+p}-1|<\epsilon$ for all $n\geq N$ and all $p\geq 1$.
Theorem: If $\prod_{k=1}^\infty a_k$ converges, then $(\prod_{k=1}^n a_k)_{n\geq 1}$ converges. If $(\prod_{k=1}^n a_k)_{n\geq 1}$ converges whose limit is non-zero then $\prod_{k=1}^\infty a_k$ converges.
If $\prod_{k=1}^\infty a_k$ converges, we write $\prod_{k=1}^\infty a_k=\lim_{n\to\infty}\prod_{k=1}^n a_k$, and the limit is called the value of the infinite product.
What I want to prove is the following properties. This is taken from the book Complex Analysis: An Invitation (2nd Edition), page 162, where the author has left the proof to the reader. This is where I need your help to see if my proofs seem correct, as I am not really good at doing $\epsilon-N$ argument.
(i) If $\prod_{k=1}^\infty a_k$ and $\prod_{k=1}^\infty b_k$ converge, then $\prod_{k=1}^\infty a_kb_k$ converges and $\prod_{k=1}^\infty a_kb_k=\left ( \prod_{k=1}^\infty a_k \right )\left ( \prod_{k=1}^\infty b_k \right )$.
(ii) If $\prod_{k=1}^\infty a_k$ converges and its value is non-zero, then $\prod_{k=1}^\infty 1/a_k$ converges and $\prod_{k=1}^\infty 1/a_k=1/\prod_{k=1}^\infty a_k$.
What I tried: (i) First, we see that $$ \left ( \prod_{k=n+1}^{n+p}a_k-1 \right )\left ( \prod_{k=n+1}^{n+p}b_k-1 \right )=\left ( \prod_{k=n+1}^{n+p} a_kb_k-1 \right )-\left ( \prod_{k=n+1}^{n+p}a_k-1 \right )-\left ( \prod_{k=n+1}^{n+p}b_k-1 \right ) $$ for all $n,p\geq 1$. Let $\epsilon\in (0,1)$ be given. Choose $N\in \mathbb{N}$ such that $|a_{n+1}\cdots a_{n+p}-1|<\epsilon$ and $|b_{n+1}\cdots b_{n+p}-1|<\epsilon$ for all $n\geq N$ and all $p\geq 1$. Then, we have $$ \left | \prod_{k=n+1}^{n+p} a_kb_k-1 \right |\leq \left | \prod_{k=n+1}^{n+p}a_k-1 \right |\left | \prod_{k=n+1}^{n+p}b_k-1 \right |+\left | \prod_{k=n+1}^{n+p}a_k-1 \right |+\left | \prod_{k=n+1}^{n+p}b_k-1 \right |<\epsilon^2+2\epsilon=\epsilon(\epsilon+2)<3\epsilon $$ for all $n\geq N$ and all $p\geq 1$, so $\prod_{k=1}^\infty a_kb_k$ converges. Moreover, we have $$ \prod_{k=1}^\infty a_kb_k=\lim_{n\to\infty}\prod_{k=1}^n a_kb_k=\lim_{n\to\infty}\left ( \prod_{k=1}^n a_k \right )\left ( \prod_{k=1}^n b_k \right )=\left ( \prod_{k=1}^\infty a_k \right )\left ( \prod_{k=1}^\infty b_k \right ) $$ Is my proof to (i) correct?
(ii) Assume $\prod_{k=1}^\infty a_k$ converges and that its value is non-zero. Put $P_n=\prod_{k=1}^n a_k$. Then, by definition, the limit $\lim_{n\to\infty}P_n$ is non-zero. Thus $\lim_{n\to\infty}1/P_n$ exists. Since $1/P_n=\prod_{k=1}^n 1/a_k$, we have by theorem, the infinite product $\prod_{k=1}^\infty 1/a_k$ converges. Its value is obvious.
