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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?

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  • $\begingroup$ Can you see that the sequence is bounded? $\endgroup$
    – saulspatz
    Commented Apr 27, 2019 at 14:04
  • $\begingroup$ Yes, I was just wanting to know where to start my answer, but there's no need for another one now. $\endgroup$
    – saulspatz
    Commented Apr 27, 2019 at 18:27
  • $\begingroup$ @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? $\endgroup$
    – UnknownW
    Commented Apr 27, 2019 at 19:39
  • $\begingroup$ 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. $\endgroup$
    – saulspatz
    Commented Apr 27, 2019 at 19:50

2 Answers 2

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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 $|C_k|<L$ 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 $C_m\not\to 0$.

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  • $\begingroup$ 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? $\endgroup$
    – UnknownW
    Commented Apr 27, 2019 at 18:41
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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 $\forall \epsilon>0$ $\forall n\geq N$ $\forall a,b\in\mathbb{N}$,

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

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    $\begingroup$ 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. $\endgroup$
    – saulspatz
    Commented Apr 27, 2019 at 14:02
  • $\begingroup$ @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 :) $\endgroup$
    – Darius
    Commented Apr 27, 2019 at 14:06

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