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Suppose $\{a_i\}_1^{\infty} \subset (0,1)$
a) $\prod_1^{\infty}(1-a_i) > 0$ iff $\sum_1^{\infty}{a_i} < \infty$
b) Given $\beta \in (0,1)$, exhibit a sequence $\{a_i\}$ such that $\prod_1^{\infty}(1-a_i) = \beta$

This is not my homework, but I'm learning measure theory from Real Analysis of Folland, and I get stuck on this problem. My idea is to prove that $\sum_1^{\infty}{\ln(1-a_i)} > -\infty$ (for sure this sum is smaller than 0). At the first glance, I try to prove that $\ln(1-x) + x > 0$, but finally, the inequality should be reversed. Using maclaurine expansion, I can expand: $$\ln(1-x) = -(x + x^2/2 + x^3/3 +...)$$

So seem like I can't find a function $f(x)$ such that $\ln(1-x) +f(x) > 0$. Can anyone give me some hint to solve this? For the second problem, I got no idea. Thanks so much. I really appreciate!

le duc quang
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  • What do you mean by $> -\infty$? – Gary. Jun 30 '15 at 02:21
  • You need to edit the title. – zhw. Jun 30 '15 at 02:28
  • @Gary: I'm not sure about the way to write it right. What I mean is to prove that that sum is finite. – le duc quang Jun 30 '15 at 02:36
  • For the forward direction, suppose $\sum_1^{\infty} a_i = \infty$. Using the fact that $1-a_i \leq e^{-a_i}$, we see that $\prod_1^n (1-a_i) \leq \prod_1^n e^{-a_i}=e^{\sum_1^n a_i}$. The latter converges to zero since $\sum a_i = \infty$. – shalin Jun 30 '15 at 02:43
  • Hint: Express the summation: $\sum_1^{\infty}{1/(1 - a_i)}$ in terms of $\prod_1^{\infty}(1-a_i)$ that would appear in the denominator of that sum. – NoChance Jun 30 '15 at 02:46
  • @leducquang: Maybe write $<< 0 $? – Gary. Jun 30 '15 at 03:03

2 Answers2

1

I think if you use $\ln(1-x)<-x$, then you can show that $% %TCIMACRO{\dprod }% %BeginExpansion {\displaystyle\prod} %EndExpansion \left( 1-a_{i}\right) >0\rightarrow% %TCIMACRO{\dsum }% %BeginExpansion {\displaystyle\sum} %EndExpansion \ln\left( 1-a_{i}\right) >-\infty\rightarrow% %TCIMACRO{\dsum }% %BeginExpansion {\displaystyle\sum} %EndExpansion -a_{i}>-\infty\rightarrow% %TCIMACRO{\dsum }% %BeginExpansion {\displaystyle\sum} %EndExpansion a_{i}<\infty.$

Now, if $% %TCIMACRO{\dsum }% %BeginExpansion {\displaystyle\sum} %EndExpansion a_{i}<\infty$, then $a_{i}\rightarrow0$. Since $\frac{\ln\left( 1-x\right) }{x}\rightarrow-1$ as $x\rightarrow0$, we have for $i$ big enough (means $a_{i}$ small enough): $\ln\left( 1-a_{i}\right) >-2a_{i}\rightarrow% %TCIMACRO{\dsum }% %BeginExpansion {\displaystyle\sum} %EndExpansion \ln\left( 1-a_{i}\right) >-2% %TCIMACRO{\dsum }% %BeginExpansion {\displaystyle\sum} %EndExpansion a_{i}>-\infty$.

For b) if you want $% %TCIMACRO{\dprod }% %BeginExpansion {\displaystyle\prod} %EndExpansion \left( 1-a_{i}\right) =\beta\rightarrow$ $% %TCIMACRO{\dsum }% %BeginExpansion {\displaystyle\sum} %EndExpansion \ln\left( 1-a_{i}\right) =\ln\beta$ then you can use the Taylor expansion to expand $\ln\beta$ and then choosing $a_{i}$ accordingly.

user72012
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0

I am just trying to add an alternative answer for the part (b) by some elementary observation.

  1. We only need to construct a sequence with $\prod_1^\infty (1-a_i)$ close to 1. For any $\beta$, if we can have $\prod_1^\infty (1-a_i)=\alpha>\beta$, then $\frac{\beta}{\alpha}<1$, then we can extend the sequence with $1-a_0= \frac{\beta}{\alpha} $ to make $\prod_0^\infty (1-a_i) = \beta$.

  2. To construct $\prod_1^\infty (1-a_i)$ close to 1, we recall the observation from a) that, when $a_i$ is small, $\sum-2a_i<\sum \ln(1-a_i)$. Pick $\beta>0, $if we can have $\ln\beta<-2\sum a_i$, then $\beta<\prod_1^\infty (1-a_i)$ as desired. But constructing $a_i$ with $\ln\beta<-2\sum a_i$ is trivial, as it only requires $-\frac{1}{2}\ln\beta>\sum a_i$, i.e the sum is small. We can have sum as small as possible as we like since we can pick $\sum_1^\infty \epsilon 2 ^{-i}= \epsilon $.

xyz
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