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I was wondering if there is any proof that the limit of infinite product

$$\lim_{n \to \infty} \prod_{i=1}^{n} x_i, \mathrm{where}$$ $$0 < x_i < 1$$

is equal to 0 and that it does not converge to 1.

I have tried to prove that power series

$$\sum_{n=1}^{\infty}\mathrm{log}(x_n)$$

does converge, hence ensuring that the product does converge to anything other than 0, but I seem to have reached a dead spot.

Thank You for any tips/ideas/solutions.

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Isn't $\prod_{i=1}^n x_i \leq x_1$? –  Hurkyl Oct 4 '12 at 18:08
    
Indeed it is, but all this inequality gives me is an information that the product is smaller than the 1st element, which in itself is saying nothing about a limit. Or am I wrong? –  Luke Brighton Oct 4 '12 at 18:39
    
By the way, for products the usual definition of "$\prod_{n=1}^\infty x_i$ converges" is not merely "$\lim_{k\to\infty}\prod_{n=1}^k x_i$ exists". One additionally requires in the definition that this limit exists when ignoring any zero terms and that this limit without zero factors is nonzero. –  Hagen von Eitzen Oct 4 '12 at 18:49
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2 Answers

up vote 2 down vote accepted

The statement $\lim_{n \to \infty} \prod_{i=1}^{n} x_i=0$ for $0 < x_i < 1$ is not true. Consider the following counterexample: let $x_i=e^{-2^{-i}}$. Then the infinite product of the $x_i$ converge to $e^{-1}$, which is non-zero.

On the other hand, the statement that the sequence does not converge to 1 is entirely true, as Hurkyl points out in the comments.

For a general treatment, recall that $0<x_i<1$ only ensures that $-\infty<\log x_i<0$, so you can have your series $\sum\log x_i$ converge to anything inside the interval $(-\infty,0)$ or diverge to $-\infty$, implying that the product converges to anything inside $e^{(-\infty,0)}=(0,1)$ or possibly converge to $0$ in the case that that $\sum\log x_i$ diverges to $-\infty$..

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Thank you very much for your explanation, it is very clear. –  Luke Brighton Oct 4 '12 at 18:55
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If you want a product which converges to a non-zero figure, take for example $x_n = 1-10^{-n} $ , i.e. $$0.9 \times 0.99 \times 0.999 \times \cdots \gt 0.89.$$

It you want a product which converges to zero even though $x_n$ tends to $1$, take for example $x_n = \dfrac{n}{n+1}$, i.e. $$\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \cdots = 0.$$

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