The root test

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.

(1) If \(\alpha < 1\) then there exists \(\epsilon > 0\) such that \(\sqrt[n]{|a_n|} < 1-\epsilon\) for sufficiently large \(n\) . This means \(|a_n| \leq (1-\epsilon)^n\) . Because \(1-\epsilon < 1\) we know the series \(\textstyle \sum (1-\epsilon)^n\) converges. According to the comparison test this implies convergence of \(\textstyle\sum a_n\).

(2) If \(\alpha > 1\) then we have \(\sqrt[n]{|a_n|}>1\) for infinitely many \(n\in\mathbb{N}\). This means that \(|a_n|>1\) for an infinite number of \(n\in\mathbb{N}\) so the terms of the series do not converge to zero. The series therefore cannot converge.

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.