Bernoulli inequality and Exponential Bound

Question

I'm asking a complete proof of the Exponential Bound using the Bernoulli Inequality.

Exponential bound:

$$1+x\leq e^x$$

Bernoulli inequality

$$1+nx\leq\left(1+x\right)^n$$ for all $$x>= -1$$ and $$n=1,2,...$$

Here it's a partial (x>=0) proof but lacks the most interesting part, when x<0.

Answer 1

Use the Bernoulli inequality with $\;\frac xn\;$ instead of $\;x\,$ , and $\;x>-1\;$ (otherwise it is very simple...) :

$$1+x\stackrel{\text{Bernoulli}}\le\left(1+\frac xn\right)^n\le e^x$$

because the central expression is a monotonic ascendent sequence whose limit is $\;e^x\;$ .

Answer 2

1.) The exponential bound is trivial for $x<-1$ so it keeps to show it for $x \ge -1$

2.) Consider that in the proof of the given link you can choose $z \ge -1$ instead of $z\ge 0$ and the proof is still valid...

And you are done.

Answer 3

Note that

• for $x<-1\implies 1+x<0 \le e^x $

• for $x>-1$

$$e\ge \left(1+\frac1x\right)^x \implies e^x\ge \left(1+\frac1x\right)^{x^2}\implies e^x\ge 1+x^2\cdot\frac1x=1+x$$