my favorite infinite series

[logistic_guy]

Solve.

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$

 

[logistic_guy]

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n} = \infty$

The series converges to infinity. I have always thought that this series diverges.

 

[khansaheb]

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n} = \infty$

The series converges to infinity. I have always thought that this series diverges.

Why? What test indicated that it is a convergent series?

 

[logistic_guy]

Why? What test indicated that it is a convergent series?

It was a joke.



One way to show that $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ diverges is to use the integral test. But before we use this test, we have to check a few things. If they are satisfied, then we can use the test.

$\displaystyle \bold{1}$ Continuous?
$\displaystyle \frac{1}{n}$ is continuous for all $\displaystyle n \geq 1$

$\displaystyle \bold{2}$ Positive?
$\displaystyle \frac{1}{n} \geq 0$ for all $\displaystyle n \geq 1$

$\displaystyle \bold{3}$ Decreasing?
$\displaystyle \frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}$
$\displaystyle \frac{1}{n}$ is decreasing for all $\displaystyle n \geq 1$


Then the theorem says:

if $\displaystyle \int_{1}^{\infty}\frac{1}{x} \ dx$ converges, then $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ converges.

Or

if $\displaystyle \int_{1}^{\infty}\frac{1}{x} \ dx$ diverges, then $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ diverges.

Using the Test.

$\displaystyle \int_{1}^{\infty}\frac{1}{x} \ dx = \ln x \bigg|_{1}^{\infty} = \ln \infty - \ln 1 = \infty - 0 = \infty$

Then, $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ diverges beautifully.