# Properties of the convergence of the infinite products

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.