# Using the integral test, how do you show whether  (1 + (1/x))^x diverges or converges?

Dec 23, 2015

it converge

#### Explanation:

By the way you can process like this

(1+1/x)^x=e^(xln(1+1/x)

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

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

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

Take the limit of $\frac{1}{1 + \frac{1}{x}}$ at $- \infty$ and $\infty$ for both case it's 1

by the squeeze theorem you can say that $\lim x \to \infty$ $x \ln \left(1 + \frac{1}{x}\right) = 1$
and $\lim x \to - \infty$ $x \ln \left(1 + \frac{1}{x}\right) = 1$

So ${\left(1 + \frac{1}{x}\right)}^{x}$ converge