Prove some inequalities of the exponential function

For $n \in \mathbb{Z}_{>0}$ and for $x \in \mathbb{R}_{>0}$ prove the following inequalities,

$\left( 1 + \frac{x}{n} \right)^n < e^x, \qquad e^x < \left( 1 - \frac{x}{n} \right)^{-n},$

where we assume $x < n$ in the second inequality. Use this to show

$2.5 < e < 2.99.$

Proof. From the previous exercise (Section 6.17, Exercise #41) we know

$e^x > 1 + x$

for $x > 0$ . But by assumption we have $x > 0$ and $n> 0$ ; hence, $\frac{x}{n} > 0$ . Therefore this inequality must hold for $\frac{x}{n}$ :

$e^{\frac{x}{n}} > 1 + \frac{x}{n} \quad \implies \quad e^x > \left( 1 + \frac{x}{n} \right)^n.$

Again, by the previous exercise we have

$e^{-x} > 1 - x$

for $x > 0$ . Since $x > 0$ and $n > 0$ this implies $\frac{x}{n} > 0$ . Therefore,

$\begin{align*} e^{-\frac{x}{n}} > 1 - \frac{x}{n} && \implies && e^{-x} > \left( 1 - \frac{x}{n} \right)^n \\ && \implies && e^x < \left( 1 - \frac{x}{n} \right)^{-n}. \qquad \blacksquare \end{align*}$

Letting $n = 6$ , we have

$\left( 1 + \frac{1}{6} \right)^6 = 2.52 < e < \left(1 - \frac{1}{6} \right)^{-6} = 2.99.$

Stumbling Robot

Prove some inequalities of the exponential function

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