# proof relating to limit definition of e

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?

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 $x$.

• 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. Mar 21, 2015 at 0:27
• As you say, it's a chiasmus. Mar 21, 2015 at 0:51

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

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

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.