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Prove some inequalities of the exponential function

by RoRi

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

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

3 comments

    • richardsull says:

      Something got messed up in my reply. I meant to say that you must have x < n or the last inequaliity is not true. The final step is valid only when x 0. If x/n > 0, then the inequality would be reversed.

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