# What is the root test?

Feb 23, 2017

Given:

${\sum}_{n = 0}^{\infty} {a}_{n}$ with ${a}_{n} \ge 0$

and:

${\lim}_{n \to \infty} \sqrt[n]{{a}_{n}} = L$

then:

$0 \le L < 1 \implies {\sum}_{n = 0}^{\infty} {a}_{n}$ is convergent

$L > 1 \implies {\sum}_{n = 0}^{\infty} {a}_{n} = \infty$

#### Explanation:

The root test states that given a series with positive terms:

${\sum}_{n = 0}^{\infty} {a}_{n}$ with ${a}_{n} \ge 0$

if the succession $\left\{\sqrt[n]{{a}_{n}}\right\}$ is convergent:

${\lim}_{n \to \infty} \sqrt[n]{{a}_{n}} = L$

then we have:

$0 \le L < 1 \implies {\sum}_{n = 0}^{\infty} {a}_{n}$ is convergent

$L > 1 \implies {\sum}_{n = 0}^{\infty} {a}_{n} = \infty$

If $L = 1$ then the test does not give us any information.

In fact, suppose that:

${\lim}_{n \to \infty} \sqrt[n]{{a}_{n}} = L < 1$

this means that for any $\epsilon > 0$ we can find $N$ such that:

$\sqrt[n]{{a}_{n}} < L + \epsilon$ for $n > N$

As $L < 1$ we can choose $\epsilon$ such that:

$L + \epsilon < 1$

Then we have, for $n > N$:

$\sqrt[n]{{a}_{n}} < L + \epsilon < 1$

and elevating both sides to the $n$-th power, which preserves the direction of the inequality:

${a}_{n} < {\left(L + \epsilon\right)}^{n}$

Now:

${\sum}_{n = 0}^{\infty} {\left(L + \epsilon\right)}^{n}$

is a geometric series of ratio $L + \epsilon < 1$ and is absolutely convergent, so also:

${\sum}_{n = 0}^{\infty} {a}_{n}$

is convergent by direct comparison.

In the same way if $L > 1$ we can establish the inequality:

${a}_{n} > {\left(L - \epsilon\right)}^{n}$ with $L - \epsilon > 1$

and determine that ${\sum}_{n = 0}^{\infty} {a}_{n}$ is divergent by direct comparison with a divergent geometric series.