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.