# Non-finite series implies product is zero

Given $0 \le y_n \le 1$ and $\sum_{n \in \mathbb{N}} y_n = \infty$, how can we show $\prod_{n=1}^\infty (1 - y_n) = 0$?

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What prevents us from taking $y_n=\frac{1}{2}\;\forall\;n\in\mathbb{N}$? – chris May 13 '12 at 5:31
Have they completely stopped teaching in schools that prefixes exist? Lately I've seen a number of different people treat "non-" as if it were a stand-alone word rather than a prefix. It's reached the point where I feel it can't be several independent typos; there must be a trend. (I corrected it in this posting.) – Michael Hardy May 13 '12 at 5:33

Note that $1-x \leq \exp(-x)$. You can verify this from calculus by looking at the function $f(x) = \exp(-x) +x - 1$ and prove that the function is increasing. Hence, $f(x) \geq f(0) = 0$. Let $M_N = \displaystyle \prod_{n=1}^{N} (1-y_n)$. Hence, we have that $$0 \leq M_N = \displaystyle \prod_{n=1}^{N} (1-y_n) \leq \displaystyle \prod_{n=1}^{N} \exp(-y_n) = \exp \left( - \sum_{n=1}^{N} y_n \right)$$ Hence, $$0 \leq \lim_{N \rightarrow \infty} M_N \leq \lim_{N \rightarrow \infty} \exp \left( - \sum_{n=1}^{N} y_n \right) \leq \lim_{N \rightarrow \infty} \frac1{1 + \displaystyle \sum_{n=1}^{N} y_n} = \lim_{N \rightarrow \infty} \frac1{1 + S(N)} = 0$$ where $\displaystyle S(N) = \sum_{n=1}^{N} y_n$ and we are given that $\displaystyle \lim_{N \rightarrow \infty} S(N) = \infty$. Hence, $$\prod_{n=1}^{\infty} \left( 1-y_n \right) = 0.$$
EDIT Since you have $0 \leq y_n \leq 1$, you could also do as follows. $$1-y_n \leq \frac1{1+y_n}.$$ As before letting, $M_N = \displaystyle \prod_{n=1}^{N} (1-y_n)$. Hence, we have that $$0 \leq M_N = \displaystyle \prod_{n=1}^{N} (1-y_n) \leq \displaystyle \prod_{n=1}^{N} \frac1{1+y_n}$$ Hence, $$0 \leq \lim_{N \rightarrow \infty} M_N \leq \lim_{N \rightarrow \infty} \prod_{n=1}^{N} \frac1{1+y_n} \leq \lim_{N \rightarrow \infty} \frac1{1 + \displaystyle \sum_{n=1}^{N} y_n} = \lim_{N \rightarrow \infty} \frac1{1 + S(N)} = 0$$ where $\displaystyle S(N) = \sum_{n=1}^{N} y_n$ and we are given that $\displaystyle \lim_{N \rightarrow \infty} S(N) = \infty$. Hence, $$\prod_{n=1}^{\infty} \left( 1-y_n \right) = 0.$$
Good answer but it could be a little bit more formal unless you want to evaluate $Exp(-\infty)$. – checkmath May 13 '12 at 5:23
@chessmath Yes. $\exp \left(- \sum_{n=1}^{\infty} y_n \right)$ was not the greatest way to write what I meant. Have changed it now. – user17762 May 13 '12 at 6:08