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I want to show that $(1+\frac{x}{n})^n \leq e^x$ for all $x \geq -n$, $n\in \mathbb{N}$.

I already showed that positive case ($x>0$) and zero case $(x=0)$.

Not sure how to approach the negative case, because in the case $x>0$ I used that $x^n$ is an increasing function on $(0, \infty)$. But I can't do the same when $x$ is negative.

letsintegreat
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Korn
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    Are you assuming $n > 0$? For $n < 0$, $(1 + \frac xn)^n$ is not always defined: for example, when $n = -\frac12$, $(1 - \frac xn)^n = \frac1{\sqrt{1-2x}}$ is not defined for any $x > \frac12$. – Misha Lavrov Feb 08 '21 at 03:19
  • Yes, $n>0$. Will edit the post now. – Korn Feb 08 '21 at 03:30

2 Answers2

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You first want to prove $1 + x \le e^x$ for all $x$, which holds because $e^x$ is convex and $1+x$ is the tangent line to $e^x$ at $(0,1)$, for example.

From $x \ge -n$ we get $\frac xn \ge -1$, or $1 + \frac xn \ge 0$. By the inequality, $1 + \frac xn \le e^{x/n}$. Because $t \mapsto t^n$ is an increasing function on $[0,\infty)$ and both sides are positive, $(1 + \frac xn)^n \le (e^{x/n})^n = e^x$.

Misha Lavrov
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Assuming you know that \begin{align} \lim_{n\rightarrow \infty}\left( 1+ \frac{x}{n}\right)^n = e^x, \end{align} then it suffices to prove that \begin{align} \left( 1+ \frac{x}{n}\right)^n \le \left( 1+ \frac{x}{n+1}\right)^{n+1} \end{align} for all $n \in \mathbb{N}$.

By Bernoulli's inequality, we have that \begin{align} \frac{\left( 1+ \frac{x}{n+1}\right)^{n+1}}{\left( 1+ \frac{x}{n}\right)^{n+1}} =&\ \left( 1+\frac{-\frac{x}{n(n+1)}}{1+\frac{x}{n}}\right)^{n+1} = \left( 1-\frac{x}{n(n+1)+(n+1)x}\right)^{n+1} \\ \ge&\ 1-(n+1)\left(\frac{x}{n(n+1)+(n+1)x} \right) = 1-\frac{x}{n+x} = \frac{1}{1+\frac{x}{n}}. \end{align} Multiplying both side by $(1+x/n)^{n+1}$ yields the desired inequality.

Remark: This is a standard proof given in a first course of analysis without assuming any knowledge of derivatives.

Jacky Chong
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