What is the root test?

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Answer 1 · 2017-02-23T13:21:51

Given:

$\infty \sum n = 0 a_{n}$ $sum_(n=0)^oo a_n$ with $a_{n} \geq 0$ $a_n >=0$

and:

$lim_(n->oo) root(n)(a_n) = L$

then:

$0 <=L < 1 => sum_(n=0)^oo a_n$ is convergent

$L > 1 => sum_(n=0)^oo a_n =oo$

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

$sum_(n=0)^oo a_n$ with $a_n >=0$

if the succession ${root(n)(a_n)}$ is convergent:

$lim_(n->oo) root(n)(a_n) = L$

then we have:

$0 <=L < 1 => sum_(n=0)^oo a_n$ is convergent

$L > 1 => sum_(n=0)^oo a_n =oo$

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

In fact, suppose that:

$lim_(n->oo) root(n)(a_n) = L < 1$

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

$root(n)(a_n) < L+epsilon$ for $n > N$

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

$L+epsilon < 1$

Then we have, for $n > N$ :

$root(n)(a_n) < L + epsilon < 1$

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

$a_n < (L+epsilon)^n$

Now:

$sum_(n=0)^oo (L+epsilon)^n$

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

$sum_(n=0)^oo a_n$

is convergent by direct comparison.

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

$a_n > (L - epsilon)^n$ with $L-epsilon > 1$

and determine that $sum_(n=0)^oo a_n$ is divergent by direct comparison with a divergent geometric series.