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I want to prove that $$ \lim_{k \to \infty} \left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k} \right) < e .$$

Using the $AM-GM$ inequality we arrive at $$\left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k} \right) < \left(\frac{k + 1 - \frac{1}{2^k} }{k} \right)^k = \left( 1 + \frac{1}{k} - \frac{1}{k2^{k}}\right)^k < \left(1 + \frac{1}{k} \right)^k < e.$$

The first inequality is strict because the terms are different.However, I know that in the limit, strict inequalities can transform into equalities. Since the limit of $\left( 1 + \frac{1}{k} - \frac{1}{k2^{k}}\right)^k$ when $k$ goes to infinity is also $e$, how could I prove a strict inequality?

mortenmcfish
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Davidmath7
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    Related: https://math.stackexchange.com/questions/1110242/inequality-prod-limits-r-1-infty-left1-frac12r-right-frac-52/2893485#2893485 – Jack D'Aurizio Jan 20 '19 at 17:49

2 Answers2

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$$ \sum_{k\geq 1}\log\left(1+\frac{1}{2^k}\right)<\sum_{k\geq 1}\frac{1}{2^k}=1,$$ then exponentiate both sides. There is also a simple strengthening, namely

$$ \sum_{k\geq 1}\log\left(1+\frac{1}{2^k}\right)<\sum_{k\geq 1}\left(\frac{1}{2^k}-\frac{1-\log 2}{4^k}\right)=\frac{2+\log 2}{3},$$ leading to $$ \prod_{k\geq 1}\left(1+\frac{1}{2^k}\right) < 2^{1/3}e^{2/3}.$$ $ \prod_{k\geq 1}\left(1+\frac{1}{2^k}\right) <\frac{31}{13}$ is a more challenging inequality.

Jack D'Aurizio
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  • How do you get that strengthening ? I guess it comes from the concavity of the log function but I don't see what value of $a$ you choose. We have : $\log(1+\frac{1}{2^k}) \leq \log(1+a) + (1+\frac{1}{2^k}-a)\frac{1}{a+1}$, but from here I don't see how to get your upper bound. – Thinking Jan 23 '19 at 08:57
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    @Thinking: I used that $\frac{x-\log(1+x)}{x^2}$ is bounded between $\frac{1}{2}$ and $1-\log 2$ for $x\in(0,1)$. – Jack D'Aurizio Jan 23 '19 at 17:08
  • Thank you ! This is actually very clever. So if I understand correctly, in order to get a sharper inequality you want to measure how the obvious upper-bound $\ln(1+x) \leq x$ isn't sharp. To do so you want to find a $k$ such that : $k \leq x-\ln(1+x)$, the problem here is that since $x \sim \ln(1+x)$ we are going to get $k = 0$. Hence to measure the speed of $x-\ln(1+x)$ we need to somehow compare it to the speed of an other function. If we divide by $x^{\alpha}, \alpha \in [1,2[$ we will get $k = 0$. So the best we can do is divide by $x^2$. Now do you think we can reiterate this method ? – Thinking Jan 23 '19 at 17:36
  • In this case we will to find the least $\alpha$ such that : $\exists k \ne 0, \forall x \in (0,1), k \leq \frac{1}{x^{\alpha}} \cdot (x-x^2(1-\log 2) - \log (1+x))$. Maybe if we continue doing so we can get something really sharp, I don't know. Don't hesitate to correct me if I am wrong. Thank you ! – Thinking Jan 23 '19 at 17:39
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    @Thinking: you are perfectly right. As shown in my other answer by carefully choosing some sorts of Padé approximants the product can be approximated with arbitrary accuracy in a reasonable time. – Jack D'Aurizio Jan 23 '19 at 18:39
  • Thank you for your answer. This is a really interesting technique. I am going to look at Padés approximant to see how I can intuitively get the same "upper-bound function" of $1+x$ in your linked answer. – Thinking Jan 23 '19 at 18:53
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You can just ignore the first term and carry on with your method from $i=2$.

Then you get $\prod\limits_{i=2}^ka_i\le\cdots\le\left(1+\dfrac 1{2(k-1)}\right)^{k-1}\le e^{0.5}$

And then $\prod\limits_{i=1}^\infty a_i=\left(1+\dfrac 12\right)\prod\limits_{i=2}^\infty a_i\le 1.5\,e^{0.5}<e$

zwim
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