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.