### Infinite series: Convergence tests

### The root test

In this section we discuss another criterion which can be used to determine whether a series converges. Just like the comparison criterion this test consists of two parts.

Let \(\textstyle\sum a_n\) be a series and define \(\alpha = \limsup \sqrt[n]{|a_n|}\) .

(1) If \(\alpha < 1\) then the series converges.

(2) If \(\alpha > 1\) then the series diverges.

For \(a_n = \frac{3+\cos(n)}{2^n}\) we have \[ \alpha = \limsup_{n\to\infty}\left|\frac{3+\cos(n)}{2^n}\right|^\frac{1}{n} = \lim_{n\to\infty}\left(\frac{4}{2^n}\right)^\frac{1}{n}=\frac{1}{2}<1\text. \] Therefore \(\textstyle \sum a_n\) converges.

Note that the root test states nothing about \(\alpha = 1\). In this case we do not obtain any information.

For \(a_n = \frac{1}{n}\) we have \(\alpha = 1\) , and for \(b_n = \frac{(-1)^n}{n}\) we have \(\alpha = 1\) as well. We have already seen that \(\sum \frac{1}{n} = \infty\) and we will see later that \(\sum\frac{(-1)^n}{n}\) indeed converges. This shows that if \(\alpha = 1\) you can't draw any conclusion about convergence of the series.