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### Section 10.11 : Root Test

This is the last test for series convergence that we’re going to be looking at. As with the Ratio Test this test will also tell whether a series is absolutely convergent or not rather than simple convergence.

#### Root Test

Suppose that we have the series \(\sum {{a_n}} \). Define,

\[L = \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{\left| {{a_n}} \right|}} = \mathop {\lim }\limits_{n \to \infty } {\left| {{a_n}} \right|^{\frac{1}{n}}}\]Then,

- if \(L < 1\) the series is absolutely convergent (and hence convergent).
- if \(L > 1\) the series is divergent.
- if \(L = 1\) the series may be divergent, conditionally convergent, or absolutely convergent.

A proof of this test is at the end of the section.

As with the ratio test, if we get \(L = 1\) the root test will tell us nothing and we’ll need to use another test to determine the convergence of the series. Also note that, generally for the series we’ll be dealing with in this class, if \(L = 1\) in the Ratio Test then the Root Test will also give \(L = 1\).

We will also need the following fact in some of these problems.

#### Fact

Let’s take a look at a couple of examples.

There really isn’t much to these problems other than computing the limit and then using the root test. Here is the limit for this problem.

\[L = \mathop {\lim }\limits_{n \to \infty } {\left| {\frac{{{n^n}}}{{{3^{1 + 2n}}}}} \right|^{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{{3^{\frac{1}{n} + 2}}}} = \frac{\infty }{{{3^2}}} = \infty > 1\]So, by the Root Test this series is divergent.

Again, there isn’t too much to this series.

\[L = \mathop {\lim }\limits_{n \to \infty } {\left| {{{\left( {\frac{{5n - 3{n^3}}}{{7{n^3} + 2}}} \right)}^n}} \right|^{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{5n - 3{n^3}}}{{7{n^3} + 2}}} \right| = \left| {\frac{{ - 3}}{7}} \right| = \frac{3}{7} < 1\]Therefore, by the Root Test this series converges absolutely and hence converges.

Note that we had to keep the absolute value bars on the fraction until we’d taken the limit to get the sign correct.

Here’s the limit for this series.

\[L = \mathop {\lim }\limits_{n \to \infty } {\left| {\frac{{{{\left( { - 12} \right)}^n}}}{n}} \right|^{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{12}}{{{n^{\frac{1}{n}}}}} = \frac{{12}}{1} = 12 > 1\]After using the fact from above we can see that the Root Test tells us that this series is divergent.

#### Proof of Root Test

First note that we can assume without loss of generality that the series will start at \(n = 1\) as we’ve done for all our series test proofs. Also note that this proof is very similar to the proof of the Ratio Test.

Let’s start off the proof here by assuming that \(L < 1\) and we’ll need to show that \(\sum {{a_n}} \) is absolutely convergent. To do this let’s first note that because \(L < 1\) there is some number \(r\) such that \(L < r < 1\).

Now, recall that,

\[L = \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{\left| {{a_n}} \right|}} = \mathop {\lim }\limits_{n \to \infty } {\left| {{a_n}} \right|^{\frac{1}{n}}}\]and because we also have chosen \(r\) such that \(L < r\) there is some \(N\) such that if \(n \ge N\) we will have,

\[{\left| {{a_n}} \right|^{\frac{1}{n}}} < r\hspace{0.5in} \Rightarrow \hspace{0.5in}\left| {{a_n}} \right| < {r^n}\]Now the series

\[\sum\limits_{n = 0}^\infty {{r^n}} \]is a geometric series and because \(0 < r < 1\) we in fact know that it is a convergent series. Also, because \(\left| {{a_n}} \right| < {r^n}\) \(n \ge N\) by the Comparison test the series

\[\sum\limits_{n = N}^\infty {\left| {{a_n}} \right|} \]is convergent. However since,

\[\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} = \sum\limits_{n = 1}^{N - 1} {\left| {{a_n}} \right|} + \sum\limits_{n = N}^\infty {\left| {{a_n}} \right|} \]we know that \(\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} \) is also convergent since the first term on the right is a finite sum of finite terms and hence finite. Therefore \(\sum\limits_{n = 1}^\infty {{a_n}} \) is absolutely convergent.

Next, we need to assume that \(L > 1\) and we’ll need to show that \(\sum {{a_n}} \) is divergent. Recalling that,

\[L = \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{\left| {{a_n}} \right|}} = \mathop {\lim }\limits_{n \to \infty } {\left| {{a_n}} \right|^{\frac{1}{n}}}\]and because \(L > 1\) we know that there must be some \(N\) such that if \(n \ge N\) we will have,

\[{\left| {{a_n}} \right|^{\frac{1}{n}}} > 1\hspace{0.5in} \Rightarrow \hspace{0.5in}\left| {{a_n}} \right| > {1^n} = 1\]However, if \(\left| {{a_n}} \right| > 1\) for all \(n \ge N\) then we know that,

\[\mathop {\lim }\limits_{n \to \infty } \left| {{a_n}} \right| \ne 0\]This in turn means that,

\[\mathop {\lim }\limits_{n \to \infty } {a_n} \ne 0\]Therefore, by the Divergence Test \(\sum {{a_n}} \) is divergent.

Finally, we need to assume that \(L = 1\) and show that we could get a series that has any of the three possibilities. To do this we just need a series for each case. We’ll leave the details of checking to you but all three of the following series have \(L = 1\) and each one exhibits one of the possibilities.

\[\begin{align*} & \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} & \hspace{0.5in} & {\mbox{absolutely convergent}}\\ & \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n}} & \hspace{0.5in} & {\mbox{conditionally convergent}}\\ & \sum\limits_{n = 1}^\infty {\frac{1}{n}} & \hspace{0.5in} & {\mbox{divergent}}\end{align*}\]