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Show that $$e^x > \bigg(1+ \frac xn\bigg)^n$$

I have thought about logarithmically deriving the sides but I cannot achieve the desired results.

Is that the correct way to go with this question?

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  • $\begingroup$ It is necessary to take $n \in \mathbb{N}$, otherwise this does not hold true. $\endgroup$ – Martigan Feb 10 '15 at 18:16
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    $\begingroup$ How do you define $e$ ? $\endgroup$ – Christian Feb 10 '15 at 18:18
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Let $x\in {\mathbb R}_{\ge 0}$. Now, note that $$ (1+\frac{x}{n})^n = \sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{x^k}{n^k}= \sum_{k=0}^n \frac{n!}{(n-k)!n^k}\frac{x^k}{k!}.$$ While this is equivalent with $$ \sum_{k=0}^n \frac{n(n-1)(n-2)...(n-k+1)}{n^k}\frac{x^k}{k!}= \sum_{k=0}^n (1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{k-1}{n})\frac{x^k}{k!}. $$ Thus $$ (1+\frac{x}{n})^n = \sum_{k=0}^n (1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{k-1}{n})\frac{x^k}{k!} \le \sum_{k=0}^n \frac{x^k}{k!} \le \sum_{k=0}^\infty \frac{x^k}{k!} =e^x. $$ For $x\in {\mathbb R}_{< 0}$, one can develop similar trend.

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There's something much simpler you can do to both sides, which enables you to express the problem in terms of $x/n$ rather than $x$.

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$$\left(\left(1+\frac{x}{n}\right)^n\right)_{n\in \mathbb N}$$ is an increasing sequence that converge to $e^x$, the result follow.

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  • $\begingroup$ It is easier to prove the inequality than the convergence of the sequence. $\endgroup$ – Martigan Feb 10 '15 at 18:17
  • $\begingroup$ The convergence is easy to prove ! $\endgroup$ – idm Feb 10 '15 at 18:48
  • $\begingroup$ @idm That does not contradict the fact that the inequality is easier to prove. $\endgroup$ – Erick Wong Feb 11 '15 at 6:30
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Raise your inequality to the power $\frac{1}{n}$ to obtain, equivalently, that $$ e^{\frac{x}{n}}>1+\frac{x}{n}. $$

Next observe, for example using the first derivative, that $0$ is a global minimum for the function $f(y)=e^y-y-1$ ($y\in\mathbb{R}$). Deduce that $f(y)\ge f(0)=0$ for every $y$ (hence also for $y=\frac{x}{n}$) with equality iff $y=0$ ($x=0$).

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Hint: $$\ln \Big(1 + \frac{x}{n}\Big) \leq \frac{x}{n} $$

For $\frac{x}{n} > -1$.

Alternatively

$f: I \to \mathbb R$ defined as $f(x) = e^x$ is convex ($f''(x) > 0 ,\forall x \in I$), then for $x \in I = (-\infty, \infty)$

$$f(x) \geq f(0) + f'(0)(x - 0) \implies e^x \geq 1 + x$$

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