show that $(1+ \frac {x}{n})^n < e^x$ and $e^x < (1- \frac{x}{n})^{-n}$ if $x<n$

Question

If $n$ is a positive integer and if $x>0$,show that

$(1+ \frac {x}{n})^n < e^x \quad$ and that $\quad e^x < (1- \frac{x}{n})^{-n} \quad $ if $x<n$

I proved the first one by the inequalities $$(1+ \frac{x}{n})^n = \sum^{n}_{k=0} \binom{n}{k} \frac{x^k}{n^k} \leq \sum_{k=0}^{n} \frac{x^k}{k!} <e^x$$

the second one is equal to $(1- \frac{x}{n})^n <e^{-x}$ but I feel I couldn't find a way to start.

In a previous exercise I proved
$$(-1)^n e^{-x} < (-1)^n \sum^{n}_{k=0}(-1)^k \frac{x^k}{k!}$$ it seems to have some relation with the second question.So I hope someone could give me a hint. Thanks in advance.

Answer 1

By convexity (or by any of a number of other approaches, for example because $\log(1+x)\leqslant x$ for every $x\gt-1$), the graph of the exponential is above its tangent at $0$, that is,

$\color{red}{\text{For every real number}\ x,}$ $$\color{red}{1+x\leqslant\mathrm e^x.}$$

Apply this to $x/n$ and raise to the power $n$, this yields the first inequality.

Apply this to $-x/n$ and raise to the power $n$, this yields $(1-x/n)^n\leqslant\mathrm e^{-x}$ hence, composing by $u\mapsto1/u$, the second inequality.