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We have the following inequality: $$e_n:=\left(1+\frac1n\right)^n<3-\frac3n,$$ for $n\in\mathbb{N}$ and $n\geq7$. My proof as follows: $e_1=2<2.75,e_2=2.25<2.75$, and for $\forall n\in\mathbb{N},n\geq2$, \begin{align*} \left(1+\frac{1}{n}\right)^n &=2+\sum_{k=2}^{n} \frac{1}{k!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\\ &\leq2+\sum_{k=2}^{n}\frac{1}{k!}\\ &\leq2+\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots+\frac{1}{3^{n-2}}\right)\\ &<2+\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots+\frac{1}{3^{n-2}}+\cdots\right)\\ &=2.75. \end{align*} When $n\in\mathbb{N}$ and $n\geq12$, $$\left(1+\frac{1}{n}\right)^n<2.75=3-\frac{3}{12}\leq3-\frac3n.$$ And for $n=7,8,9,10,11$, we can check the inequality directly (or use wolfram mathematica).

What I want to konw: Are there any elementary proofs of this inequality? Any help and hints will welcome!

Riemann
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1 Answers1

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This isn't exactly elementary but does the job. We wish to show that $f(x)<3$ for all real $x\ge7$ where $f(x)=(1+1/x)^x/(1-1/x)$. Its derivative is given by $$f'(x)=g(x)\left(x(x^2-1)\log\left(1+1/x\right)-x^2-1\right)$$ where $g(x)=(1+1/x)^x/((x-1)^2(x+1))$ is positive. Since $$x(x^2-1)\log(1+1/x)-x^2-1<(x^2-1)(x\log(1+1/x)-1)<0$$ for all $x>1$, we know that $f(x)$ is strictly decreasing for all $x>1$. The claim follows as $f(7)<3$.

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