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I'm studying the book A Primer for the Mathematics of FinancialEngineering, along with it's solution guide. I've ran into trouble with the solution to question 7 in chapter 1.

The problem is:

Show that $(1+\frac{1}{x})^x < e < (1+\frac{1}{x})^{x+1}$ .

Specifically, I'm having trouble with one step in the solution.

It says that

$$x\ln\left(1+\frac{1}{x}\right)<1<(x+1)\ln\left(1+\frac{1}{x}\right)$$

can be written as

$$\frac{1}{x+1} <\ln\left(1+\frac{1}{x}\right)<\frac{1}{x}, \forall \ge 1.$$

How can the first equality (or is it inequality?) be written as the second?

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3 Answers 3

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The inequality $1<(x+1)\ln(1+\frac{1}{x})$ becomes $\frac{1}{x+1} <\ln(1+\frac{1}{x})$ upon dividing by $x+1$, and the inequality $x\ln(1+\frac{1}{x})<1$ becomes $\ln(1+\frac{1}{x})<\frac{1}{x}$ upon dividing by $x$.

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    $\begingroup$ In other words, the left hand inequality becomes the right hand inequality (and vice versa), and you have to split the inequality into two to make that happen. $\endgroup$
    – Teepeemm
    Mar 21, 2015 at 0:27
  • $\begingroup$ As you say, it's a chiasmus. $\endgroup$ Mar 21, 2015 at 0:51
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Express the log function as $\log (x) = \int_1^x \frac{du}{u}$, and note that this is the inverse of the exponential function $e^x$.

Inasmuch as the integrand is an increasing function, it is straightforward to show that

$$\frac{1}{x+1}=\frac{1}{1+1/x}\left((1+1/x)-1\right)<\int_1^{1+1/x} \frac{1}{u}\,du < (1)\left((1+1/x)-1\right)=\frac{1}{x}$$

Now, exponentiation of terms reveals

$$e^{1/(x+1)} < 1+\frac{1}{x} < e^{1/x}$$

from which the desired inequalities emerge as

$$\left(1+\frac{1}{x}\right)^x < e < \left(1+\frac{1}{x}\right)^{x+1}$$

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you can easily show that $x\mapsto (1+\frac{1}{x})^x$ is increasing, $x\mapsto (1+\frac{1}{x})^{x+1}$ is decreasing and that both converge to $e$. Therefore you got your inequality.

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