# About the definition of an infinite product and its analysis

Definition: Let $$c_1,c_2,\dots$$ be a sequence of complex numbers. The infinite product $$\prod_{k=1}^{\infty}c_k$$ is said to converge if there to each $$\epsilon>0$$ exists an $$N\in\mathbb{N}$$ such that for all $$n\geq N$$ and $$p\geq 1$$ $$\left | \prod_{k=n+1}^{n+p}c_k-1 \right |<\epsilon \tag{1}$$

It's mentioned in the book page 162:

If the infinite product $$\prod_{k=1}^{\infty}c_k$$ converges, then the numbers $$C_n:=\prod_{k=1}^{n}c_k$$, $$n=1,2,\dots$$ form a bounded sequence. Using (1) we find that $$(C_n)$$ is a Cauchy sequence and therefore has a limit in $$\mathbb{C}$$. We use the notation $$\prod_{k=1}^{\infty}c_k$$ also for this limit and say that it is the value of the infinite product.

and

Also note that if $$\prod_{k=1}^{\infty}c_k$$ is a convergent product, then $$\prod_{k=1}^{\infty}c_k=0$$ if and only if at least one of the factors $$c_k$$ is zero.

Could anyone explain why $$(C_n)$$ is a Cauchy sequence, and why is the last quotation true?

• Can you see that the sequence is bounded? Commented Apr 27, 2019 at 14:04
• Yes, I was just wanting to know where to start my answer, but there's no need for another one now. Commented Apr 27, 2019 at 18:27
• @saulspatz That's fine. I'm struggling to do the last part of this post. Let $\prod_{k=1}^\infty c_k$ be convergent. If $c_k=0$ for some $k$, then $C_n=0$ when $n\geq k$ and so $\prod_{j=1}^\infty c_j=0$ (not sure about this one). Assume conversely $\prod_{k=1}^\infty c_k=0$. Then $(C_n)$ converges to $0$. Therefore for all $\epsilon >0$, there exists $N$ such that $\prod_{k=1}^{N}|c_k|=|C_N-0|<\epsilon$. Since $\epsilon$ is arbitrary, we get $\prod_{k=1}^{N}|c_k|=0$, which implies that $|c_k|=0$ for some $k=1,2,\dots, N$. This completes the proof. Is this correct, or am I missing something? Commented Apr 27, 2019 at 19:39
• The second part of this is wrong. $N$ depends on $\varepsilon$. You are resonng as if it were the same $N$ for every $\varepsilon.$ See Hagen von Eitzen's answer for a poof of this. Commented Apr 27, 2019 at 19:50

If $$C_k=0$$ for some $$k$$ then $$C_k=0$$ for all $$k>n$$ and the sequence is clearly Cauchy. So assume $$C_k\ne 0$$ for all $$k$$. As we know that the sequence $$C_n$$ is bounded, we can pick $$L>0$$ with $$|C_k| for all $$k$$. Fix $$\epsilon>0$$. By assumption, for $$\epsilon':=\frac {\epsilon}L$$ there exists $$N$$ such that $$n>N$$ and $$p\ge 1$$, we have $$\left|\prod _{k=n+1}^{n+p} c_k-1\right|<\epsilon'$$

Equivalently, for all $$n,m>N$$, where wlog $$m>n$$, $$\left|\frac {C_m}{C_n}-1\right|<\frac\epsilon L.$$ Thus for such $$n,m$$ $$|C_m-C_n|<\frac{\epsilon|C_n|}{L}<\epsilon.$$

If the product converges, then for $$\epsilon=\frac12$$ there exists $$N$$ such that $$\left|\prod_{k=n+1}^m-1\right|<\frac12$$ for all $$m>n>N$$. In particular, either $$C_{N+1}=0$$, or $$\left|\frac{C_m}{C_{N+1}}-1\right|<\frac12$$, whence $$|C_m|>\frac12 |C_{n+1}|$$. Hence either $$C_{N+1}=0$$ or $$C_m\not\to 0$$.

• I do not understand the last part, "In particular, either $C_{N+1}=0$, or $\left|\frac{C_m}{C_{N+1}}-1\right|<\frac12$, whence $|C_m|>\frac12 |C_{n+1}|$". Where does $C_{N+1}=0$ come from? Commented Apr 27, 2019 at 18:41

Could anyone explain why $$(C_n)$$ is a Cauchy sequence?

All convergent sequences are Cauchy sequences. In particular,

Since the sequence converges, we can choose $$\epsilon>0$$ such that whenever $$n\geq N_1$$ $$\forall a\in\mathbb{N}$$, $$\left|\prod_{k=n+1}^{n+p+a}c_k-1 \right|<\frac{\epsilon}{2}$$.

Similarly, we can choose $$\epsilon>0$$ such that whenever $$n\geq N_2\forall b\in\mathbb{N}$$, $$\left|\prod_{k=n+1}^{n+p+b}c_k-1 \right|<\frac{\epsilon}{2}$$.

So let $$N=\max(\{N_1,N_2\})$$. Then $$\forall \epsilon>0$$ $$\forall n\geq N$$ $$\forall a,b\in\mathbb{N}$$,

\begin{align*}\left|\prod_{k=n+1}^{n+p+a}c_k - \prod_{k=n+1}^{n+p+b}c_k\right|&=\left|\prod_{k=n+1}^{n+p+a}c_k - \prod_{k=n+1}^{n+p+b}c_k-1 +1\right|\\&<\left|\prod_{k=n+1}^{n+p+b}c_k-1 \right| + \left|\prod_{k=n+1}^{n+p+a}c_k-1 \right|\\ &< \frac{\epsilon}{2}+\frac{\epsilon}{2}\\&=\epsilon\end{align*}.

• If I'm not mistaken, this doesn't prove that it's a Cauchy sequence. It's not enough to prove that the difference between two consecutive terms is arbitrarily small. Commented Apr 27, 2019 at 14:02
• @saulspatz, oh yes you're right! But it's simple to modify the argument so that it fits but my main point was always that any convergent sequence is a Cauchy sequence :) Commented Apr 27, 2019 at 14:06