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.
On
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. $$\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \cdots = 0.$$
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 $-\infty$..