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I want to show that $\prod_{n=1}^{\infty}(1-x_{n}) > 0$ if $x_{n} \in [0, 1)$ and $\sum_{n=1}^{\infty}x_{n} < \infty$.

So far I used the inequality $1 + t \leq e^{t}$ for all $t\in\mathbb{R}$ to bound $\prod_{n=1}^{\infty}(1-x_{n}) \leq e^{M}$ for some $M$. Then I just quicky used the Monotone Convergence Theorem as the product is montone decreasing to conclude that the product converges. However, I am having troubles showing that it does not converge to $0$ and therefore $\prod_{n=1}^{\infty}(1-x_{n}) > 0$. Any help is appreciated.

Tomislav
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If $\ \displaystyle\sum_{n=1}^\infty x_n<\infty\ $, then $\ \lim_\limits{n\rightarrow\infty}x_n=0\ $, so there is a positive integer $\ N\ $ such that $\ x_n\le\frac{1}{2}\ $ for $ n\ge N\ $. Then $$ 1-x_n\ge e^{-2x_n}\ \text{ for }\ n\ge N. $$ Therefore, \begin{align} \prod_{n=N}^{N+m}\left(1-x_n\right)&\ge e^{-2\sum_{n=N}^{N+m}x_n}\\ &\ge e^{-2\sum_{n=N}^\infty x_n}\ , \end{align} and so \begin{align} \prod_{n=N}^\infty\left(1-x_n\right)&\ge e^{-2\sum_{n=N}^\infty x_n}\\ &>0\ . \end{align} Since $\ \prod_\limits{n=1}^{N-1}\left(1-x_n\right)>0\ $, it follows that $\ \prod_\limits{n=1}^\infty\left(1-x_n\right)>0\ $.

lonza leggiera
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  • I really liked this proof. In particular how you broke the product up into two pieces: finite and infinite part. How did you come up with $1-x_n\ge e^{-2x_n}\ \text{ for }\ n\ge N$. This was the most useful part. – Tomislav Sep 30 '20 at 19:49
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    I had a vague recollection that the result could be proved by taking logarithms, although I don't recall whether I have ever seen such a proof. On writing down the inequality $$ \ln(1−x)\ge−\frac{x}{1−x} $$ I realised that the tail of $\ \sum_n\ln(1−x_n)\ $ could be bounded below by that of $\ −K\sum_nx_n\ $ for some constant $\ K\ $. Splitting the product isn't really necessary, though. If $\ K=\frac{1}{1−\sup_n x_n}\ $, then $$ \prod_\limits{n=1}^\infty\left(1-x_n\right)> e^{-K\sum_{n=1}^\infty x_n}\ . $$ – lonza leggiera Oct 01 '20 at 01:47
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$$\sum x_n < +\infty \Longrightarrow \lim_{n \rightarrow +\infty} x_n = 0$$

so $$-\ln(1-x_n) \sim x_n$$

so by comparison, the series $\sum \ln(1-x_n)$ converges to a limit $L$, so $$\prod_{n=1}^N (1-x_n) = \exp \left( \sum_{n=1}^N \ln(1-x_n)\right)$$

converges when $N$ tends to $+\infty$ to $$\prod_{n=1}^{+\infty} (1-x_n) = \exp(L) >0$$

TheSilverDoe
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