If I could show that than it would be equivalent to say that there is no natural number in the interval $ (2,(1+\frac{1}{n})^{n})] \Rightarrow (1+\frac{1}{n})^{n} < 3 $
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I want to show what's on the right side of the implication but binommialcoefficients or logarithmus or roots were not introduced yet. That's because I am looking for a very specific solution to this problem. Maybe itcan be solvedby using the method of supremum and infinum. – RM777 Nov 06 '18 at 23:09
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You have already asked how to prove $e<3$ without binomial coefficients, and $\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n<1$ is not equivalent to $e<3$. – Jack D'Aurizio Nov 06 '18 at 23:09
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Additionally, I have already answered you that you may prove $e<3$ (actually $e<\frac{20}{7}$) by telescopic products only, page 106 here. – Jack D'Aurizio Nov 06 '18 at 23:11
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Can you please tell me which pages of the book do I have to read in order to understand the proof? All the pages prior to the proof? – RM777 Nov 06 '18 at 23:15
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No one, the proof is a one-liner, once you know that $\left(1+\frac{1}{n}\right)^n$ is increasing. – Jack D'Aurizio Nov 06 '18 at 23:18
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Hint:
$\Bigl(1+\dfrac1{n+1}\Bigr)^{\!n+1}<\mathrm e$. On the other hand, $\Bigl(1+\dfrac1n\Bigr)^{\!n}>1+n\cdot\dfrac1n=2$ (Bernoulli's inequality), so $$\Bigl(1+\dfrac1{n+1}\Bigr)^{\!n+1}-\Bigl(1+\dfrac1n\Bigr)^{\!n}<\mathrm e-2.$$
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Alternatively you can use the fact that $(1 + 1/n)^n$ is increasing to show that $(1+1/n)^n \geq (1+1)^1 = 2$. – Contravariant Nov 06 '18 at 23:44
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