The following is from Tom Apostol's Calculus I, on page 250, exercise 42.:

If $\mathit{n}$ is a positive integer and if $\mathit{x} > 0$, show that
$$ \left( 1 + \frac{x}{n} \right)^n < e^x \text, \qquad \text{and that} \qquad e^x < \left(1 - \frac{x}{n}\right)^{-n} \quad \text{if} \quad x < n. $$

There is no solution provided in the book, I would like to ask someone to verify if mine is correct:

Integrating the inequality $ 1 > \frac{1}{1 + \frac{x}{n}} $ we get that

$$ \int_0^x{1dt} > \int_0^x{\frac{1}{1 + \frac{t}{n}}dt} \qquad \text{yielding} \qquad x > n\log\left(1 + \frac{x}{n}\right). $$

Since $ e $ is strictly increasing it follows that $$ e^x > \left(1 + \frac{x}{n}\right)^{n}. $$

Similarly, integrating $ 1 < \frac{1}{1 - \frac{x}{n}} $ we get that $$ \int_0^x{1dt} < \int_0^x{\frac{1}{1 - \frac{t}{n}}dt} \qquad \text{yielding} \qquad x < -n\log\left(1 - \frac{x}{n}\right) . $$ It follows that $$ e^x <\left(1 - \frac{x}{n}\right)^{-n} . $$

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    $\begingroup$ Simpler is to note that $e^x\ge1+x$ for all $x\in\mathbb{R}$. However, both your derivations look fine. $\endgroup$ – robjohn Oct 5 '15 at 20:14
  • $\begingroup$ Thanks for going through it. Could you explain how we get from your inequality to the ones in the question? $\endgroup$ – Imre Deák Oct 5 '15 at 20:24
  • $\begingroup$ I have added an answer with some more detail. $\endgroup$ – robjohn Oct 5 '15 at 20:29
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    $\begingroup$ If you look at Exercise 41 in Apostol, you'll see it asks you to deduce the inequality robjohn cites. $\endgroup$ – Barry Cipra Oct 5 '15 at 20:53
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    $\begingroup$ @ImreDeák, it's always good to step back and look for what you might have overlooked, but actually I find a lot to like in the approach you took to proving the inequalities. $\endgroup$ – Barry Cipra Oct 7 '15 at 20:22

Your proofs look fine.

Another approach is to use the fact that $e^x\ge1+x$ for all $x\in\mathbb{R}$, with strict inequality for $x\ne0$. This follows from the strict convexity of $e^x-1-x$ and its minimum of $0$ at $x=0$.

Then we have for $x\gt0$, $$ e^{x/n}\gt1+\frac xn\implies e^x\gt\left(1+\frac xn\right)^n\tag{1} $$ and $$ e^{-x/n}\gt1-\frac xn\implies e^x\lt\left(1-\frac xn\right)^{-n}\tag{2} $$ Inequality $(2)$ does rely on the fact that $x\lt n$ so that the quantity being raised to the negative power is positive.


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